360268
Statement A : In a mixture of gases at a fixed temperature, the heavier molecule has the lower average speed. Statement B : Temperature of a gas is a measure of the average kinetic energy of a molecule.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\(\dfrac{1}{2} m v_{\text {avg }}^{2}=\sqrt{\dfrac{8 R T}{\pi M}} \Rightarrow v_{\text {avg }} \propto \dfrac{1}{m}\) So statement A is correct.\(K.{\rm{ }}{E_{avg{\rm{ }}}} \propto T\) Statement B is also correct. So option (3) is correct.
PHXI13:KINETIC THEORY
360269
Choose the only corret statement from the following:
1 The pressure of a gas is equal to the total kinetic energy of the molecules in a unit volume of the gas.
2 The product of pressure and volume of a gas is always constant.
3 The average kinetic energy of molecules of a gas is proportional to its absolute temperature.
4 The average kinetic energy of molecules of a gas is proportional to the square root of its absolute temperature.
Explanation:
Conceptual Question
PHXI13:KINETIC THEORY
360270
The temperature of a gas is \({-78^{\circ} {C}}\) and the average translational kinetic energy of its molecules is \({K}\). The temperature at which the average translational kinetic energy of the molecules of the same gas becomes \({2 K}\) is
1 \({127^{\circ} {C}}\)
2 \({-39^{\circ} {C}}\)
3 \({117^{\circ} {C}}\)
4 \({-78^{\circ} {C}}\)
Explanation:
\({T_{i}=-78^{\circ} {C}=195 {~K}}\) As we know that, \(K.E. \propto T,\) when kinetic energy is doubled, temperature also double. \({K_{f}=2 k \Rightarrow T_{f}=2 T}\) \({T_{f}=2 \times 195=390 {~K}=117^{\circ} {C}}\) So, correct option is (3).
JEE - 2024
PHXI13:KINETIC THEORY
360271
A light container having a diatomic gas enclosed with in is moving with velocity \(v\). Mass of the gas is \(M\) and number of moles is \(n\). The kinetic energy of the gas w.r.to ground is
1 \(\dfrac{1}{2} M v^{2}\)
2 \(\dfrac{5}{2} n R T\)
3 \(\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
4 \(\dfrac{1}{2} n R T\)
Explanation:
The kinetic energy of gas w.r.t. centre of mass of the system is the kinetic energy of molecules due to their random motion. This will be the internal energy of the gas. \(\mathrm{K} . \mathrm{E} .=\dfrac{5}{2} n R T\) Given that Mass of gas \( = M\), Temperature \( = T\) Kinetic energy of gas w.r.t. ground \(=\) Kinetic energy of centre of mass w.r.t. ground + Kinetic energy of gas w.r.t. centre of mass. \(K . E=\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
PHXI13:KINETIC THEORY
360272
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature \({T}\). Neglecting all vibrational modes, the total internal energy of the system is
1 \({20 R T}\)
2 \({29 R T}\)
3 \(27\,RT\)
4 \({21 R T}\)
Explanation:
Number of moles of argon, \({n_{1}=8}\) Number of moles of oxygen, \({n_{2}=6}\) For an ideal gas, internal energy, \({U=\dfrac{f}{2} n R T}\) where ' \({f}\) ' is degree of freedom \({\therefore U_{1}=\dfrac{f_{1}}{2} n_{1} R T}\) and \({U_{2}=\dfrac{f_{2}}{2} n_{2} R T}\) Total energy, \({U=U_{1}+U_{2}}\) \({U=\dfrac{f_{1}}{2} n_{1} R T+\dfrac{f_{2}}{2} n_{2} R T}\) \({U=\dfrac{3}{2} \times 8 R T+\dfrac{5}{2} \times 6 R T=27 R T}\) So, correct option is (3).
360268
Statement A : In a mixture of gases at a fixed temperature, the heavier molecule has the lower average speed. Statement B : Temperature of a gas is a measure of the average kinetic energy of a molecule.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\(\dfrac{1}{2} m v_{\text {avg }}^{2}=\sqrt{\dfrac{8 R T}{\pi M}} \Rightarrow v_{\text {avg }} \propto \dfrac{1}{m}\) So statement A is correct.\(K.{\rm{ }}{E_{avg{\rm{ }}}} \propto T\) Statement B is also correct. So option (3) is correct.
PHXI13:KINETIC THEORY
360269
Choose the only corret statement from the following:
1 The pressure of a gas is equal to the total kinetic energy of the molecules in a unit volume of the gas.
2 The product of pressure and volume of a gas is always constant.
3 The average kinetic energy of molecules of a gas is proportional to its absolute temperature.
4 The average kinetic energy of molecules of a gas is proportional to the square root of its absolute temperature.
Explanation:
Conceptual Question
PHXI13:KINETIC THEORY
360270
The temperature of a gas is \({-78^{\circ} {C}}\) and the average translational kinetic energy of its molecules is \({K}\). The temperature at which the average translational kinetic energy of the molecules of the same gas becomes \({2 K}\) is
1 \({127^{\circ} {C}}\)
2 \({-39^{\circ} {C}}\)
3 \({117^{\circ} {C}}\)
4 \({-78^{\circ} {C}}\)
Explanation:
\({T_{i}=-78^{\circ} {C}=195 {~K}}\) As we know that, \(K.E. \propto T,\) when kinetic energy is doubled, temperature also double. \({K_{f}=2 k \Rightarrow T_{f}=2 T}\) \({T_{f}=2 \times 195=390 {~K}=117^{\circ} {C}}\) So, correct option is (3).
JEE - 2024
PHXI13:KINETIC THEORY
360271
A light container having a diatomic gas enclosed with in is moving with velocity \(v\). Mass of the gas is \(M\) and number of moles is \(n\). The kinetic energy of the gas w.r.to ground is
1 \(\dfrac{1}{2} M v^{2}\)
2 \(\dfrac{5}{2} n R T\)
3 \(\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
4 \(\dfrac{1}{2} n R T\)
Explanation:
The kinetic energy of gas w.r.t. centre of mass of the system is the kinetic energy of molecules due to their random motion. This will be the internal energy of the gas. \(\mathrm{K} . \mathrm{E} .=\dfrac{5}{2} n R T\) Given that Mass of gas \( = M\), Temperature \( = T\) Kinetic energy of gas w.r.t. ground \(=\) Kinetic energy of centre of mass w.r.t. ground + Kinetic energy of gas w.r.t. centre of mass. \(K . E=\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
PHXI13:KINETIC THEORY
360272
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature \({T}\). Neglecting all vibrational modes, the total internal energy of the system is
1 \({20 R T}\)
2 \({29 R T}\)
3 \(27\,RT\)
4 \({21 R T}\)
Explanation:
Number of moles of argon, \({n_{1}=8}\) Number of moles of oxygen, \({n_{2}=6}\) For an ideal gas, internal energy, \({U=\dfrac{f}{2} n R T}\) where ' \({f}\) ' is degree of freedom \({\therefore U_{1}=\dfrac{f_{1}}{2} n_{1} R T}\) and \({U_{2}=\dfrac{f_{2}}{2} n_{2} R T}\) Total energy, \({U=U_{1}+U_{2}}\) \({U=\dfrac{f_{1}}{2} n_{1} R T+\dfrac{f_{2}}{2} n_{2} R T}\) \({U=\dfrac{3}{2} \times 8 R T+\dfrac{5}{2} \times 6 R T=27 R T}\) So, correct option is (3).
360268
Statement A : In a mixture of gases at a fixed temperature, the heavier molecule has the lower average speed. Statement B : Temperature of a gas is a measure of the average kinetic energy of a molecule.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\(\dfrac{1}{2} m v_{\text {avg }}^{2}=\sqrt{\dfrac{8 R T}{\pi M}} \Rightarrow v_{\text {avg }} \propto \dfrac{1}{m}\) So statement A is correct.\(K.{\rm{ }}{E_{avg{\rm{ }}}} \propto T\) Statement B is also correct. So option (3) is correct.
PHXI13:KINETIC THEORY
360269
Choose the only corret statement from the following:
1 The pressure of a gas is equal to the total kinetic energy of the molecules in a unit volume of the gas.
2 The product of pressure and volume of a gas is always constant.
3 The average kinetic energy of molecules of a gas is proportional to its absolute temperature.
4 The average kinetic energy of molecules of a gas is proportional to the square root of its absolute temperature.
Explanation:
Conceptual Question
PHXI13:KINETIC THEORY
360270
The temperature of a gas is \({-78^{\circ} {C}}\) and the average translational kinetic energy of its molecules is \({K}\). The temperature at which the average translational kinetic energy of the molecules of the same gas becomes \({2 K}\) is
1 \({127^{\circ} {C}}\)
2 \({-39^{\circ} {C}}\)
3 \({117^{\circ} {C}}\)
4 \({-78^{\circ} {C}}\)
Explanation:
\({T_{i}=-78^{\circ} {C}=195 {~K}}\) As we know that, \(K.E. \propto T,\) when kinetic energy is doubled, temperature also double. \({K_{f}=2 k \Rightarrow T_{f}=2 T}\) \({T_{f}=2 \times 195=390 {~K}=117^{\circ} {C}}\) So, correct option is (3).
JEE - 2024
PHXI13:KINETIC THEORY
360271
A light container having a diatomic gas enclosed with in is moving with velocity \(v\). Mass of the gas is \(M\) and number of moles is \(n\). The kinetic energy of the gas w.r.to ground is
1 \(\dfrac{1}{2} M v^{2}\)
2 \(\dfrac{5}{2} n R T\)
3 \(\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
4 \(\dfrac{1}{2} n R T\)
Explanation:
The kinetic energy of gas w.r.t. centre of mass of the system is the kinetic energy of molecules due to their random motion. This will be the internal energy of the gas. \(\mathrm{K} . \mathrm{E} .=\dfrac{5}{2} n R T\) Given that Mass of gas \( = M\), Temperature \( = T\) Kinetic energy of gas w.r.t. ground \(=\) Kinetic energy of centre of mass w.r.t. ground + Kinetic energy of gas w.r.t. centre of mass. \(K . E=\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
PHXI13:KINETIC THEORY
360272
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature \({T}\). Neglecting all vibrational modes, the total internal energy of the system is
1 \({20 R T}\)
2 \({29 R T}\)
3 \(27\,RT\)
4 \({21 R T}\)
Explanation:
Number of moles of argon, \({n_{1}=8}\) Number of moles of oxygen, \({n_{2}=6}\) For an ideal gas, internal energy, \({U=\dfrac{f}{2} n R T}\) where ' \({f}\) ' is degree of freedom \({\therefore U_{1}=\dfrac{f_{1}}{2} n_{1} R T}\) and \({U_{2}=\dfrac{f_{2}}{2} n_{2} R T}\) Total energy, \({U=U_{1}+U_{2}}\) \({U=\dfrac{f_{1}}{2} n_{1} R T+\dfrac{f_{2}}{2} n_{2} R T}\) \({U=\dfrac{3}{2} \times 8 R T+\dfrac{5}{2} \times 6 R T=27 R T}\) So, correct option is (3).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI13:KINETIC THEORY
360268
Statement A : In a mixture of gases at a fixed temperature, the heavier molecule has the lower average speed. Statement B : Temperature of a gas is a measure of the average kinetic energy of a molecule.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\(\dfrac{1}{2} m v_{\text {avg }}^{2}=\sqrt{\dfrac{8 R T}{\pi M}} \Rightarrow v_{\text {avg }} \propto \dfrac{1}{m}\) So statement A is correct.\(K.{\rm{ }}{E_{avg{\rm{ }}}} \propto T\) Statement B is also correct. So option (3) is correct.
PHXI13:KINETIC THEORY
360269
Choose the only corret statement from the following:
1 The pressure of a gas is equal to the total kinetic energy of the molecules in a unit volume of the gas.
2 The product of pressure and volume of a gas is always constant.
3 The average kinetic energy of molecules of a gas is proportional to its absolute temperature.
4 The average kinetic energy of molecules of a gas is proportional to the square root of its absolute temperature.
Explanation:
Conceptual Question
PHXI13:KINETIC THEORY
360270
The temperature of a gas is \({-78^{\circ} {C}}\) and the average translational kinetic energy of its molecules is \({K}\). The temperature at which the average translational kinetic energy of the molecules of the same gas becomes \({2 K}\) is
1 \({127^{\circ} {C}}\)
2 \({-39^{\circ} {C}}\)
3 \({117^{\circ} {C}}\)
4 \({-78^{\circ} {C}}\)
Explanation:
\({T_{i}=-78^{\circ} {C}=195 {~K}}\) As we know that, \(K.E. \propto T,\) when kinetic energy is doubled, temperature also double. \({K_{f}=2 k \Rightarrow T_{f}=2 T}\) \({T_{f}=2 \times 195=390 {~K}=117^{\circ} {C}}\) So, correct option is (3).
JEE - 2024
PHXI13:KINETIC THEORY
360271
A light container having a diatomic gas enclosed with in is moving with velocity \(v\). Mass of the gas is \(M\) and number of moles is \(n\). The kinetic energy of the gas w.r.to ground is
1 \(\dfrac{1}{2} M v^{2}\)
2 \(\dfrac{5}{2} n R T\)
3 \(\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
4 \(\dfrac{1}{2} n R T\)
Explanation:
The kinetic energy of gas w.r.t. centre of mass of the system is the kinetic energy of molecules due to their random motion. This will be the internal energy of the gas. \(\mathrm{K} . \mathrm{E} .=\dfrac{5}{2} n R T\) Given that Mass of gas \( = M\), Temperature \( = T\) Kinetic energy of gas w.r.t. ground \(=\) Kinetic energy of centre of mass w.r.t. ground + Kinetic energy of gas w.r.t. centre of mass. \(K . E=\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
PHXI13:KINETIC THEORY
360272
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature \({T}\). Neglecting all vibrational modes, the total internal energy of the system is
1 \({20 R T}\)
2 \({29 R T}\)
3 \(27\,RT\)
4 \({21 R T}\)
Explanation:
Number of moles of argon, \({n_{1}=8}\) Number of moles of oxygen, \({n_{2}=6}\) For an ideal gas, internal energy, \({U=\dfrac{f}{2} n R T}\) where ' \({f}\) ' is degree of freedom \({\therefore U_{1}=\dfrac{f_{1}}{2} n_{1} R T}\) and \({U_{2}=\dfrac{f_{2}}{2} n_{2} R T}\) Total energy, \({U=U_{1}+U_{2}}\) \({U=\dfrac{f_{1}}{2} n_{1} R T+\dfrac{f_{2}}{2} n_{2} R T}\) \({U=\dfrac{3}{2} \times 8 R T+\dfrac{5}{2} \times 6 R T=27 R T}\) So, correct option is (3).
360268
Statement A : In a mixture of gases at a fixed temperature, the heavier molecule has the lower average speed. Statement B : Temperature of a gas is a measure of the average kinetic energy of a molecule.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\(\dfrac{1}{2} m v_{\text {avg }}^{2}=\sqrt{\dfrac{8 R T}{\pi M}} \Rightarrow v_{\text {avg }} \propto \dfrac{1}{m}\) So statement A is correct.\(K.{\rm{ }}{E_{avg{\rm{ }}}} \propto T\) Statement B is also correct. So option (3) is correct.
PHXI13:KINETIC THEORY
360269
Choose the only corret statement from the following:
1 The pressure of a gas is equal to the total kinetic energy of the molecules in a unit volume of the gas.
2 The product of pressure and volume of a gas is always constant.
3 The average kinetic energy of molecules of a gas is proportional to its absolute temperature.
4 The average kinetic energy of molecules of a gas is proportional to the square root of its absolute temperature.
Explanation:
Conceptual Question
PHXI13:KINETIC THEORY
360270
The temperature of a gas is \({-78^{\circ} {C}}\) and the average translational kinetic energy of its molecules is \({K}\). The temperature at which the average translational kinetic energy of the molecules of the same gas becomes \({2 K}\) is
1 \({127^{\circ} {C}}\)
2 \({-39^{\circ} {C}}\)
3 \({117^{\circ} {C}}\)
4 \({-78^{\circ} {C}}\)
Explanation:
\({T_{i}=-78^{\circ} {C}=195 {~K}}\) As we know that, \(K.E. \propto T,\) when kinetic energy is doubled, temperature also double. \({K_{f}=2 k \Rightarrow T_{f}=2 T}\) \({T_{f}=2 \times 195=390 {~K}=117^{\circ} {C}}\) So, correct option is (3).
JEE - 2024
PHXI13:KINETIC THEORY
360271
A light container having a diatomic gas enclosed with in is moving with velocity \(v\). Mass of the gas is \(M\) and number of moles is \(n\). The kinetic energy of the gas w.r.to ground is
1 \(\dfrac{1}{2} M v^{2}\)
2 \(\dfrac{5}{2} n R T\)
3 \(\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
4 \(\dfrac{1}{2} n R T\)
Explanation:
The kinetic energy of gas w.r.t. centre of mass of the system is the kinetic energy of molecules due to their random motion. This will be the internal energy of the gas. \(\mathrm{K} . \mathrm{E} .=\dfrac{5}{2} n R T\) Given that Mass of gas \( = M\), Temperature \( = T\) Kinetic energy of gas w.r.t. ground \(=\) Kinetic energy of centre of mass w.r.t. ground + Kinetic energy of gas w.r.t. centre of mass. \(K . E=\dfrac{1}{2} M v^{2}+\dfrac{5}{2} n R T\)
PHXI13:KINETIC THEORY
360272
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature \({T}\). Neglecting all vibrational modes, the total internal energy of the system is
1 \({20 R T}\)
2 \({29 R T}\)
3 \(27\,RT\)
4 \({21 R T}\)
Explanation:
Number of moles of argon, \({n_{1}=8}\) Number of moles of oxygen, \({n_{2}=6}\) For an ideal gas, internal energy, \({U=\dfrac{f}{2} n R T}\) where ' \({f}\) ' is degree of freedom \({\therefore U_{1}=\dfrac{f_{1}}{2} n_{1} R T}\) and \({U_{2}=\dfrac{f_{2}}{2} n_{2} R T}\) Total energy, \({U=U_{1}+U_{2}}\) \({U=\dfrac{f_{1}}{2} n_{1} R T+\dfrac{f_{2}}{2} n_{2} R T}\) \({U=\dfrac{3}{2} \times 8 R T+\dfrac{5}{2} \times 6 R T=27 R T}\) So, correct option is (3).