360260
If the molecular weight of two gases are \(M_{1}\) and \(M_{2}\), then at a temperature the ratio of rms velocity \(c_{1}\) and \(c_{2}\) will be
As, \(\dfrac{1}{2} M c^{2}=\dfrac{3}{2} R T\) or \(c=\left(\dfrac{3 R T}{M}\right)^{1 / 2}\) \(\Rightarrow c \propto \dfrac{1}{\sqrt{M}}\) So, \(\quad \dfrac{c_{1}}{c_{2}}=\left(\dfrac{M_{2}}{M_{1}}\right)^{1 / 2}\)
PHXI13:KINETIC THEORY
360261
Assertion : The root mean square and most probable speeds of the molecules in a gas are the same. Reason : The Maxwell distribution for the speed of molecules in a gas is symmetrical.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect.
Explanation:
According to kinetic theory of gases molecules do exhibit random motion, hence assertion is incorrect. Also Maxwell's velocity distribution is asymmetric. So correct option is (4).
AIIMS - 2006
PHXI13:KINETIC THEORY
360262
The ratio of the densities of two gases at the same temperature is \(8: 9\). The ratio of the rms velocities of their molecules is
1 \(1: 1\)
2 \(8: 9\)
3 \(\sqrt{8}: \sqrt{9}\)
4 \(\sqrt{9}: \sqrt{8}\)
Explanation:
RMS velocity depends only on temperature. So the ratio of rms velocities is \(1: 1\)
PHXI13:KINETIC THEORY
360263
Calculate the rms speed of smoke particles each of mass \(5 \times {10^{ - 17}} kg\) in thier Brownian motion in air at NTP \(\left(k=1.38 \times 10^{-23} J K^{-1}\right)\)
360260
If the molecular weight of two gases are \(M_{1}\) and \(M_{2}\), then at a temperature the ratio of rms velocity \(c_{1}\) and \(c_{2}\) will be
As, \(\dfrac{1}{2} M c^{2}=\dfrac{3}{2} R T\) or \(c=\left(\dfrac{3 R T}{M}\right)^{1 / 2}\) \(\Rightarrow c \propto \dfrac{1}{\sqrt{M}}\) So, \(\quad \dfrac{c_{1}}{c_{2}}=\left(\dfrac{M_{2}}{M_{1}}\right)^{1 / 2}\)
PHXI13:KINETIC THEORY
360261
Assertion : The root mean square and most probable speeds of the molecules in a gas are the same. Reason : The Maxwell distribution for the speed of molecules in a gas is symmetrical.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect.
Explanation:
According to kinetic theory of gases molecules do exhibit random motion, hence assertion is incorrect. Also Maxwell's velocity distribution is asymmetric. So correct option is (4).
AIIMS - 2006
PHXI13:KINETIC THEORY
360262
The ratio of the densities of two gases at the same temperature is \(8: 9\). The ratio of the rms velocities of their molecules is
1 \(1: 1\)
2 \(8: 9\)
3 \(\sqrt{8}: \sqrt{9}\)
4 \(\sqrt{9}: \sqrt{8}\)
Explanation:
RMS velocity depends only on temperature. So the ratio of rms velocities is \(1: 1\)
PHXI13:KINETIC THEORY
360263
Calculate the rms speed of smoke particles each of mass \(5 \times {10^{ - 17}} kg\) in thier Brownian motion in air at NTP \(\left(k=1.38 \times 10^{-23} J K^{-1}\right)\)
360260
If the molecular weight of two gases are \(M_{1}\) and \(M_{2}\), then at a temperature the ratio of rms velocity \(c_{1}\) and \(c_{2}\) will be
As, \(\dfrac{1}{2} M c^{2}=\dfrac{3}{2} R T\) or \(c=\left(\dfrac{3 R T}{M}\right)^{1 / 2}\) \(\Rightarrow c \propto \dfrac{1}{\sqrt{M}}\) So, \(\quad \dfrac{c_{1}}{c_{2}}=\left(\dfrac{M_{2}}{M_{1}}\right)^{1 / 2}\)
PHXI13:KINETIC THEORY
360261
Assertion : The root mean square and most probable speeds of the molecules in a gas are the same. Reason : The Maxwell distribution for the speed of molecules in a gas is symmetrical.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect.
Explanation:
According to kinetic theory of gases molecules do exhibit random motion, hence assertion is incorrect. Also Maxwell's velocity distribution is asymmetric. So correct option is (4).
AIIMS - 2006
PHXI13:KINETIC THEORY
360262
The ratio of the densities of two gases at the same temperature is \(8: 9\). The ratio of the rms velocities of their molecules is
1 \(1: 1\)
2 \(8: 9\)
3 \(\sqrt{8}: \sqrt{9}\)
4 \(\sqrt{9}: \sqrt{8}\)
Explanation:
RMS velocity depends only on temperature. So the ratio of rms velocities is \(1: 1\)
PHXI13:KINETIC THEORY
360263
Calculate the rms speed of smoke particles each of mass \(5 \times {10^{ - 17}} kg\) in thier Brownian motion in air at NTP \(\left(k=1.38 \times 10^{-23} J K^{-1}\right)\)
360260
If the molecular weight of two gases are \(M_{1}\) and \(M_{2}\), then at a temperature the ratio of rms velocity \(c_{1}\) and \(c_{2}\) will be
As, \(\dfrac{1}{2} M c^{2}=\dfrac{3}{2} R T\) or \(c=\left(\dfrac{3 R T}{M}\right)^{1 / 2}\) \(\Rightarrow c \propto \dfrac{1}{\sqrt{M}}\) So, \(\quad \dfrac{c_{1}}{c_{2}}=\left(\dfrac{M_{2}}{M_{1}}\right)^{1 / 2}\)
PHXI13:KINETIC THEORY
360261
Assertion : The root mean square and most probable speeds of the molecules in a gas are the same. Reason : The Maxwell distribution for the speed of molecules in a gas is symmetrical.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect.
Explanation:
According to kinetic theory of gases molecules do exhibit random motion, hence assertion is incorrect. Also Maxwell's velocity distribution is asymmetric. So correct option is (4).
AIIMS - 2006
PHXI13:KINETIC THEORY
360262
The ratio of the densities of two gases at the same temperature is \(8: 9\). The ratio of the rms velocities of their molecules is
1 \(1: 1\)
2 \(8: 9\)
3 \(\sqrt{8}: \sqrt{9}\)
4 \(\sqrt{9}: \sqrt{8}\)
Explanation:
RMS velocity depends only on temperature. So the ratio of rms velocities is \(1: 1\)
PHXI13:KINETIC THEORY
360263
Calculate the rms speed of smoke particles each of mass \(5 \times {10^{ - 17}} kg\) in thier Brownian motion in air at NTP \(\left(k=1.38 \times 10^{-23} J K^{-1}\right)\)
