189. At what temperature degree Fahrenheit scale shows twice the reading of degree centigrade? [ 160 0C] Hints: The relation between C and F scale of temperature is . Take C = , then F = 2.
190. At what temperature degree Fahrenheit scale shows half of the reading of degree centigrade? [ – 24.62 0C] Hints: Take C = , then F will be /2 and put the value in relation
191. At what temperature Fahrenheit and Kelvin scale give the same reading?
[ 574.25K]
Hints: Take the temperature F = , then K is also and put the value in relation
192. The normal temperature of human body is 37 0C. What is its value in 0F and Kelvin scale? [ 98.60F, 310 K]
193. A faulty thermometer has its fixed points –100C and 130 0C. This thermometer reads the temperature of an object 60 0C. Find the correct temperature of the body in Celsius scale. [ 500 C ] Hints: The term where L is lower fix point and U is upper fix point, remains constant for all scale of thermometer. Use same expression for both correct thermometer and faulty thermometer.
194. A faulty Celsius thermometer reads – 4 0C when placed in melting ice and reads 98 0C when placed in the contact with steam at normal pressure. What the correct temperature in that scale when it reads room temperature 32 0C ? [ 35.3 0C ]
190. At what temperature degree Fahrenheit scale shows half of the reading of degree centigrade? [ – 24.62 0C] Hints: Take C = , then F will be /2 and put the value in relation
191. At what temperature Fahrenheit and Kelvin scale give the same reading?
[ 574.25K]
Hints: Take the temperature F = , then K is also and put the value in relation
192. The normal temperature of human body is 37 0C. What is its value in 0F and Kelvin scale? [ 98.60F, 310 K]
193. A faulty thermometer has its fixed points –100C and 130 0C. This thermometer reads the temperature of an object 60 0C. Find the correct temperature of the body in Celsius scale. [ 500 C ] Hints: The term where L is lower fix point and U is upper fix point, remains constant for all scale of thermometer. Use same expression for both correct thermometer and faulty thermometer.
194. A faulty Celsius thermometer reads – 4 0C when placed in melting ice and reads 98 0C when placed in the contact with steam at normal pressure. What the correct temperature in that scale when it reads room temperature 32 0C ? [ 35.3 0C ]
Thermal Expansion
195. A steel meter scale has length 100.00 cm at temperature 10 0C. If the temperature rises to 20 0C, what is the length of the scale? ( for steel 1.6 x 10-6 / k). If a road of length 1 km is measured at 20 0C, what will be the error in the measurement? Hints: The length of the scale at temperature 20 0C is given by l2 = l1 [ 1+ (2 – 1)]. Divide length of the road by the length of the scale at 20 0C to obtain its reading.
196. An aluminum rod when measured with a steel scale, both being at 25 0C appears to be 1 m long. If the scale is correct at 0 0C, (a) what is the true length of the scale at that temperature? (b) What will be the length of the rod at 0 0C? Linear expansivity of aluminum is 26 x 20-6 /K and of steel is 12 x 10-6/K. [1.0003m, 99.96m] Hints:
197. A glass vessel of volume 200 cm3 is just filled with mercury at 100C. If the temperature raised to 120 0C, how much mercury overflow? ( for glass 1.2 x 10-6 / k and for mercury 1.6 x 10 –4 /k ) [ 3.44 cm3] Hints: The overflowing mercury is the difference between the volume of mercury and the volume of vessel at 120 0C. So, determine volume of mercury and volume of vessel at 120 0C separately and calculate the difference between them.
198. A thermal tap used in certain apparatus consists a silica rod which fits tightly inside an Aluminum tube whose internal diameter is 8 mm at 00 C. When temperature is raised, the fit is no longer exact. Calculate what change of temperature is necessary to provide a channel whose cross-section is equal to that of a tube of 1mm internal diameter. Linear expansivity of Si and Al are 8 x 10-6/K and 26 x 10-6 /K. [ 434 K ]
Hints: Let at temperature , the channel area will be equal to the cross section area of a tube of 1 mm internal diameter which is given by . The channel area is equal to the difference between the cross-section areas of the cylinder(Al) to piston(Si). So, (A)Al – (A)Si = .
195. A steel meter scale has length 100.00 cm at temperature 10 0C. If the temperature rises to 20 0C, what is the length of the scale? ( for steel 1.6 x 10-6 / k). If a road of length 1 km is measured at 20 0C, what will be the error in the measurement? Hints: The length of the scale at temperature 20 0C is given by l2 = l1 [ 1+ (2 – 1)]. Divide length of the road by the length of the scale at 20 0C to obtain its reading.
196. An aluminum rod when measured with a steel scale, both being at 25 0C appears to be 1 m long. If the scale is correct at 0 0C, (a) what is the true length of the scale at that temperature? (b) What will be the length of the rod at 0 0C? Linear expansivity of aluminum is 26 x 20-6 /K and of steel is 12 x 10-6/K. [1.0003m, 99.96m] Hints:
197. A glass vessel of volume 200 cm3 is just filled with mercury at 100C. If the temperature raised to 120 0C, how much mercury overflow? ( for glass 1.2 x 10-6 / k and for mercury 1.6 x 10 –4 /k ) [ 3.44 cm3] Hints: The overflowing mercury is the difference between the volume of mercury and the volume of vessel at 120 0C. So, determine volume of mercury and volume of vessel at 120 0C separately and calculate the difference between them.
198. A thermal tap used in certain apparatus consists a silica rod which fits tightly inside an Aluminum tube whose internal diameter is 8 mm at 00 C. When temperature is raised, the fit is no longer exact. Calculate what change of temperature is necessary to provide a channel whose cross-section is equal to that of a tube of 1mm internal diameter. Linear expansivity of Si and Al are 8 x 10-6/K and 26 x 10-6 /K. [ 434 K ]
Hints: Let at temperature , the channel area will be equal to the cross section area of a tube of 1 mm internal diameter which is given by . The channel area is equal to the difference between the cross-section areas of the cylinder(Al) to piston(Si). So, (A)Al – (A)Si = .
199. A brass pendulum clock gives correct time at 15 0C. How many second does it lose or gain per day when the temperature changes to 20 0C? ( for brass 0.000019 / k ) [ 4 s ]
Hints: Time period of pendulum at 15 0C is and that of 20 0C is . Since correct time of pendulum is 2 second, put T1 = 2 second. The relation between l2 and l1 is . Put the value of l2 and divide T1 to T2. Determine difference between T1 and T2 and finally determine gain or lose in time per day.
200. A steel wire 8m long and 4 mm in diameter is fixed to two rigid support. Calculate the increase in tension when the temperature falls by 100 C? Given linear expansivity of steel 12 x 10-6 /K, Young modulus for steel 2 x 1011 N/m2 . [ 300 N ]
Hints: Young’s modulus Y = , where F is force or tension, l is initial length, A is cross section area and e is extension or contraction of the wire. Use relation or Or, e = where e = is the contraction. Finally use the relation of young’s modulus to calculate tension.
201. A steel cylinder has an Aluminum piston and at temperature of 200 C, internal diameter of cylinder is exactly 10 cm, there is an around clearance of 0.05 mm between the piston and the cylinder wall. At what temperature will the fit be perfect? Given linear expansivity of steel and Aluminum are 1.2 x 10-5 /K and 1.6 x 10-5 /K respectively. [ 2700 C ]
Hints: To fit perfectly, the diameter of steel piston should be exactly equal to the diameter of aluminum cylinder. Consider the temperature at which their diameters are exactly equal. So, .
Hints: Time period of pendulum at 15 0C is and that of 20 0C is . Since correct time of pendulum is 2 second, put T1 = 2 second. The relation between l2 and l1 is . Put the value of l2 and divide T1 to T2. Determine difference between T1 and T2 and finally determine gain or lose in time per day.
200. A steel wire 8m long and 4 mm in diameter is fixed to two rigid support. Calculate the increase in tension when the temperature falls by 100 C? Given linear expansivity of steel 12 x 10-6 /K, Young modulus for steel 2 x 1011 N/m2 . [ 300 N ]
Hints: Young’s modulus Y = , where F is force or tension, l is initial length, A is cross section area and e is extension or contraction of the wire. Use relation or Or, e = where e = is the contraction. Finally use the relation of young’s modulus to calculate tension.
201. A steel cylinder has an Aluminum piston and at temperature of 200 C, internal diameter of cylinder is exactly 10 cm, there is an around clearance of 0.05 mm between the piston and the cylinder wall. At what temperature will the fit be perfect? Given linear expansivity of steel and Aluminum are 1.2 x 10-5 /K and 1.6 x 10-5 /K respectively. [ 2700 C ]
Hints: To fit perfectly, the diameter of steel piston should be exactly equal to the diameter of aluminum cylinder. Consider the temperature at which their diameters are exactly equal. So, .
202. Aniline is a liquid which doesn’t mix with water and when a small quantity of it is poured into beaker of water at 200 C, it sinks to the bottom, the densities of two liquids at 200 C are 1021 and 998 Kg/m3 respectively. At what temperature must the beaker and its contents be heated so that the aniline will form a globule which just floats in the water? Given absolute expansivity of aniline and water are 0.00085 /K and 0.00045 /K respectively. [ 790 C ]
Hints: To satisfy the mentioned condition, density of aniline must be equal to the density of water. So, let at temperature , their densities are exactly equal i.e. .
Hints: To satisfy the mentioned condition, density of aniline must be equal to the density of water. So, let at temperature , their densities are exactly equal i.e. .
203. A certain Fortin barometer has its pointer, body and scale made from brass. When it is at 00 C, it records pressure 760 mmHg. What will it read when its temperature is increased to 200C if pressure of the atmospheric remains unchanged? Given cubical expansivity of mercury 1.8 x 10-4 /K, Linear expansivity of brass 2 x 10-5 /K. [ 762.4 mm Hg ] Hints: Use the expression of correction of barometer.
204. Using the following data, determine the temperature at which wood will just sink in benzene? Density of benzene at 00C = 900 kg / m3. Density of wood at 00C = 880 kg / m3. Cubical expansivity of benzene = 0.0012 / k
Cubical expansivity of wood = 0.0015 / k
Hints: To satisfy the mentioned condition, density of wood should be exactly equal to the density of the benzene. So, let at temperature , their densities are exactly equal i.e.
Calorimetry / Change of state
205. A steel ball of mass 80 gram is taken from a furnace of temperature 2000C and dropped in a copper calorimeter of mass 50 gram containing 65 gm of water at temperature 20 0C. If the final temperature of the mixture rises to 22.5 0C, calculate the specific heat capacity of the steel. ( Specific heat capacity of copper is 0.1 cal / gm k. Hints: In the above phenomenon, steel ball loss heat but calorimeter and water gains heat. According to principle of calorimetry, heat loss is equal to the heat gain. So, heat loss by steel ball Q1 = m S Heat gain by caloriemeter and water Q2 = (m S ) for water + (m S ) for calorimeter. From principle of calorimetry, Q1 = Q2
206. A copper ball of mass 15 g is held in a flame until it has acquired the temperature of the flame. It is then quickly transferred to a copper calorimeter of mass 66 g containing 50 g of water at 20 0C. If the final steady temperature of the mixture is 27.5 0C, calculate the temperature of the flame. [323 0C]
207. Two identical calorimeters each of heat capacity 12 J/K, one contains 8 x 10-5 m3 of water and takes 150 second to cool from 325 K to 320 K, and the other contains an equal volume of an unknown liquid which takes 50 second to cool over the same range of temperature. If the density of the liquid is 800 kg/m3, what is the specific heat capacity of liquid over the temperature range? Given , density of water 1000 kg/m3 and heat capacity 4200 J/K. [ 1625 L/kgK]
208. A lead bullet moving with velocity of 400m/s strikes a target and is brought to rest. If half of the kinetic energy goes to raise the temperature of the bullet, by how many degrees will its temperature rise? Given specific heat capacity of the bullet is 0.13 J/g 0C . [ 307.20C] Hints: According to question, half of the KE changes into heat. So, ( ½ mv2 ) = m x s x
209. A substance takes 3 min in cooling from 50 0C to 45 0C and takes 5 min in cooling from 45 0C to 40 0C. What is the temperature of its surrounding? How much time will it take to cool from 40 0C to 35 0C ? [ 35 0C, 15 min ]
210. How much energy is required to convert 500 g ice of temperature –5 0C to steam of temperature 100 0C ? ( Specific and latent heat of ice are 0.5 cal /gm 0C and 80 cal /gm and latent heat of steam is 340 cal / gm. )
211. 5 gm of ice at 0 0C is dropped into a beaker containing 20 g of water at 40 0C. What will be the final temperature? [ 16 0C ]
212. What is the result of mixture when 400g of water and 100 g of ice at 00C are in a copper calorimeter of mass 500g and 100 g of steam is passed into it? [
213. What is the result of mixing 20g of ice at -10 0C and 50g of water at 300C ? [ Specific heat of ice 2100 J/kg K & latent heat of ice 3.34 x 105 J/kg ]
214. 1 g of steam at 100 0C can melt how much ice at 0 0C ? [ 8 g ]
Ideal Gas Equation
215. A container of gas has volume 0.1 m3 at a pressure of 2 x 105 N/m2 and a temperature of 27 0C. (a) Find the new pressure if the gas is heated at constant volume to 87 0C, (b) The gas pressure is now reduced to 1.0 x 105 N/m2 at constant temperature. What is the new volume of the gas? (c) The gas cooled to – 73 0C at constant pressure. Find the new volume of the gas. [2.4 x 105 N/m2, 0.24 m3, 0.13 m3 ]
216. A cylinder of gas has mass of 10.0 kg and pressure of 8.0 atmosphere at 27 0C. When some gas is used in a cooled room at – 3 0C, the gas remaining in the cylinder at this temperature has a pressure 6.4 atm. Calculate the mass of the gas used. [ 1.1 kg ] Hints: From ideal gas equation, P1V = m1 r T1 ………..(i) When some gas is used in a cool room, the equation for the remaining gas will be P2V = m2 r T2 ………..(ii) Dividing equation (i) by (ii),
217. Two glass bulb of equal volume are joined by a narrow tube and are filled with a gas at stp. When one bulb is kept in melting ice and the other is placed in hot bath, the new pressure is 877.6 mm of Hg. Calculate the temperature of the bath. [ 100 0C ] Hints: When both bulbs are in STP, the ideal gas equation will be P1 (2V) = m r T1 So, m = When one bulb is placed in melting ice, P1 V = m1 r T1 So, m1 = When another bulb is placed in hot bath, mass of gas in the bulb is m2 = Since mass of gas remains constant, m1 + m2 = m
218. Two vessels of capacity 1.00 litre are connected by a tube of negligible volume. Together, they contain 3.42 x 10-4 kg of helium at a pressure of 800 mm of mercury at temperature 27 0C. Calculate (a) value for the constant ‘r’ for helium, (b) the pressure developed in the apparatus if one vessel is cooled to 0 0C and next is heated to 100 0C. [ 2080 J/kg K, 842 mm ]
Hints: Both bulbs are in same pressure, the ideal gas equation will be P (2V) = m r T where, pressure P = 800 mm of Hg = 800 x 13600×9.8 Nm2 volume V = 1 litre = 10-3 m3 mass m = 3.42 x 10-4 Kg temperature T = 270C = 300 K
219. What volume of liquid oxygen( density 1140 kg/m3) may be made by liquefying completely the contents of a cylinder of gaseous oxygen containing 100 litre at 120 atm pressure at 20 0C? Assume that oxygen behaves as an ideal gas in this latter region of pressure and temperature. Given 1 atmosphere = 1.01 x 105 N/m2, molar gas constant is 8.31 J /molK and relative molecular mass of oxygen is 32. [ 0.014 m3 ]
220. A sealed bottle full of water is placed in a strong container full of air at standard atmospheric pressure and at a temperature of 10 0C. The temperature in the container is raised to and maintained at 100 0C. Neglecting the expansion of the bottle and the container, what is the new pressure in the container? If the bottle breaks, what will be the pressure be? [ 1.3 x 105 N/m2, 2.3 x 105]
221. A container of gas has a volume of 0. 1 m3 at pressure of 2 x 105 N / m2 and temperature 27 0C. (i) Find the new pressure if the gas is heated at constant volume to 87 0C. (ii) The pressure is now reduced to 1 atmosphere at constant temperature.
Kinetic theory of gas
222. Calculate the root – mean- square speed at 0 0C of (i) Hydrogen molecules and (ii) Oxygen molecules, assuming 1 mole of the gas occupies a volume of 2 x 10-2 m3 at 0 0C and pressure 105 N/m2. Relative molecular masses of hydrogen and oxygen are 2 and 32 respectively. [1732 m/s, 433 m/s]
Hints: The pressure exerted by gas in terms of root mean square speed is given by . Replace density by mass per unit volume and finally determine C.
223. Assuming helium molecules have a root mean square speed of 900 m/s at 27 0C and 105 pressure, calculate the rms speed at (i) 127 0C and 105 N/m2, (ii) 27 0C and 2 x 105 N/m2 pressure. [ 1039 m/s, 900 m/s]
Hints:
204. Using the following data, determine the temperature at which wood will just sink in benzene? Density of benzene at 00C = 900 kg / m3. Density of wood at 00C = 880 kg / m3. Cubical expansivity of benzene = 0.0012 / k
Cubical expansivity of wood = 0.0015 / k
Hints: To satisfy the mentioned condition, density of wood should be exactly equal to the density of the benzene. So, let at temperature , their densities are exactly equal i.e.
Calorimetry / Change of state
205. A steel ball of mass 80 gram is taken from a furnace of temperature 2000C and dropped in a copper calorimeter of mass 50 gram containing 65 gm of water at temperature 20 0C. If the final temperature of the mixture rises to 22.5 0C, calculate the specific heat capacity of the steel. ( Specific heat capacity of copper is 0.1 cal / gm k. Hints: In the above phenomenon, steel ball loss heat but calorimeter and water gains heat. According to principle of calorimetry, heat loss is equal to the heat gain. So, heat loss by steel ball Q1 = m S Heat gain by caloriemeter and water Q2 = (m S ) for water + (m S ) for calorimeter. From principle of calorimetry, Q1 = Q2
206. A copper ball of mass 15 g is held in a flame until it has acquired the temperature of the flame. It is then quickly transferred to a copper calorimeter of mass 66 g containing 50 g of water at 20 0C. If the final steady temperature of the mixture is 27.5 0C, calculate the temperature of the flame. [323 0C]
207. Two identical calorimeters each of heat capacity 12 J/K, one contains 8 x 10-5 m3 of water and takes 150 second to cool from 325 K to 320 K, and the other contains an equal volume of an unknown liquid which takes 50 second to cool over the same range of temperature. If the density of the liquid is 800 kg/m3, what is the specific heat capacity of liquid over the temperature range? Given , density of water 1000 kg/m3 and heat capacity 4200 J/K. [ 1625 L/kgK]
208. A lead bullet moving with velocity of 400m/s strikes a target and is brought to rest. If half of the kinetic energy goes to raise the temperature of the bullet, by how many degrees will its temperature rise? Given specific heat capacity of the bullet is 0.13 J/g 0C . [ 307.20C] Hints: According to question, half of the KE changes into heat. So, ( ½ mv2 ) = m x s x
209. A substance takes 3 min in cooling from 50 0C to 45 0C and takes 5 min in cooling from 45 0C to 40 0C. What is the temperature of its surrounding? How much time will it take to cool from 40 0C to 35 0C ? [ 35 0C, 15 min ]
210. How much energy is required to convert 500 g ice of temperature –5 0C to steam of temperature 100 0C ? ( Specific and latent heat of ice are 0.5 cal /gm 0C and 80 cal /gm and latent heat of steam is 340 cal / gm. )
211. 5 gm of ice at 0 0C is dropped into a beaker containing 20 g of water at 40 0C. What will be the final temperature? [ 16 0C ]
212. What is the result of mixture when 400g of water and 100 g of ice at 00C are in a copper calorimeter of mass 500g and 100 g of steam is passed into it? [
213. What is the result of mixing 20g of ice at -10 0C and 50g of water at 300C ? [ Specific heat of ice 2100 J/kg K & latent heat of ice 3.34 x 105 J/kg ]
214. 1 g of steam at 100 0C can melt how much ice at 0 0C ? [ 8 g ]
Ideal Gas Equation
215. A container of gas has volume 0.1 m3 at a pressure of 2 x 105 N/m2 and a temperature of 27 0C. (a) Find the new pressure if the gas is heated at constant volume to 87 0C, (b) The gas pressure is now reduced to 1.0 x 105 N/m2 at constant temperature. What is the new volume of the gas? (c) The gas cooled to – 73 0C at constant pressure. Find the new volume of the gas. [2.4 x 105 N/m2, 0.24 m3, 0.13 m3 ]
216. A cylinder of gas has mass of 10.0 kg and pressure of 8.0 atmosphere at 27 0C. When some gas is used in a cooled room at – 3 0C, the gas remaining in the cylinder at this temperature has a pressure 6.4 atm. Calculate the mass of the gas used. [ 1.1 kg ] Hints: From ideal gas equation, P1V = m1 r T1 ………..(i) When some gas is used in a cool room, the equation for the remaining gas will be P2V = m2 r T2 ………..(ii) Dividing equation (i) by (ii),
217. Two glass bulb of equal volume are joined by a narrow tube and are filled with a gas at stp. When one bulb is kept in melting ice and the other is placed in hot bath, the new pressure is 877.6 mm of Hg. Calculate the temperature of the bath. [ 100 0C ] Hints: When both bulbs are in STP, the ideal gas equation will be P1 (2V) = m r T1 So, m = When one bulb is placed in melting ice, P1 V = m1 r T1 So, m1 = When another bulb is placed in hot bath, mass of gas in the bulb is m2 = Since mass of gas remains constant, m1 + m2 = m
218. Two vessels of capacity 1.00 litre are connected by a tube of negligible volume. Together, they contain 3.42 x 10-4 kg of helium at a pressure of 800 mm of mercury at temperature 27 0C. Calculate (a) value for the constant ‘r’ for helium, (b) the pressure developed in the apparatus if one vessel is cooled to 0 0C and next is heated to 100 0C. [ 2080 J/kg K, 842 mm ]
Hints: Both bulbs are in same pressure, the ideal gas equation will be P (2V) = m r T where, pressure P = 800 mm of Hg = 800 x 13600×9.8 Nm2 volume V = 1 litre = 10-3 m3 mass m = 3.42 x 10-4 Kg temperature T = 270C = 300 K
219. What volume of liquid oxygen( density 1140 kg/m3) may be made by liquefying completely the contents of a cylinder of gaseous oxygen containing 100 litre at 120 atm pressure at 20 0C? Assume that oxygen behaves as an ideal gas in this latter region of pressure and temperature. Given 1 atmosphere = 1.01 x 105 N/m2, molar gas constant is 8.31 J /molK and relative molecular mass of oxygen is 32. [ 0.014 m3 ]
220. A sealed bottle full of water is placed in a strong container full of air at standard atmospheric pressure and at a temperature of 10 0C. The temperature in the container is raised to and maintained at 100 0C. Neglecting the expansion of the bottle and the container, what is the new pressure in the container? If the bottle breaks, what will be the pressure be? [ 1.3 x 105 N/m2, 2.3 x 105]
221. A container of gas has a volume of 0. 1 m3 at pressure of 2 x 105 N / m2 and temperature 27 0C. (i) Find the new pressure if the gas is heated at constant volume to 87 0C. (ii) The pressure is now reduced to 1 atmosphere at constant temperature.
Kinetic theory of gas
222. Calculate the root – mean- square speed at 0 0C of (i) Hydrogen molecules and (ii) Oxygen molecules, assuming 1 mole of the gas occupies a volume of 2 x 10-2 m3 at 0 0C and pressure 105 N/m2. Relative molecular masses of hydrogen and oxygen are 2 and 32 respectively. [1732 m/s, 433 m/s]
Hints: The pressure exerted by gas in terms of root mean square speed is given by . Replace density by mass per unit volume and finally determine C.
223. Assuming helium molecules have a root mean square speed of 900 m/s at 27 0C and 105 pressure, calculate the rms speed at (i) 127 0C and 105 N/m2, (ii) 27 0C and 2 x 105 N/m2 pressure. [ 1039 m/s, 900 m/s]
Hints:
224. Air may be taken to consist of 80 % nitrogen molecules and 20 %oxygen molecules of relative molecular masses 28 and 32 respectively. Calculate (a) ratio of the rms speed of nitrogen molecules to that of oxygen molecules, (b) ratio of the partial pressure of nitrogen and oxygen molecules, (c) ratio of the rms speed of nitrogen molecules at 100C to that at 2000C. [ 1.07:1, 4:1, 0.87:1]
225. Calculate the pressure in mm of Hg exerted by Hydrogen gas if the number of molecules per cm3 is 6.80 x 1015 and the root mean square speed of the molecules is 1.90 x 103 m/s. Given Avogadro constant is 6.02 x 1023 and relative molecular mass of hydrogen is 2.02. [ 0.21 mm of Hg ]
226. Assuming that the density of Nitrogen at stp to be 1.251 kg /m3 find the root mean square velocity of nitrogen molecules at 127 0C. [ 597 m/s ]
Hints: At first determine rms speed of nitrogen at stp and then determine its value at another temperature by using relation .
227. Air at 273 K and 1.01 x 105 N/m2 pressure contains 2.7 x 1025 molecules per cubic meter. How many molecules per cubic meter will there be at a place where temperature is 223 K and pressure is 1.33 x 104 N / m2
225. Calculate the pressure in mm of Hg exerted by Hydrogen gas if the number of molecules per cm3 is 6.80 x 1015 and the root mean square speed of the molecules is 1.90 x 103 m/s. Given Avogadro constant is 6.02 x 1023 and relative molecular mass of hydrogen is 2.02. [ 0.21 mm of Hg ]
226. Assuming that the density of Nitrogen at stp to be 1.251 kg /m3 find the root mean square velocity of nitrogen molecules at 127 0C. [ 597 m/s ]
Hints: At first determine rms speed of nitrogen at stp and then determine its value at another temperature by using relation .
227. Air at 273 K and 1.01 x 105 N/m2 pressure contains 2.7 x 1025 molecules per cubic meter. How many molecules per cubic meter will there be at a place where temperature is 223 K and pressure is 1.33 x 104 N / m2
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