139275
The rms velocity of a gas at $T^{\circ} \mathrm{C}$ is double the value at $27^{\circ} \mathrm{C}$. The temperature $\mathrm{T}$ of the gas in ${ }^{0} \mathrm{C}$ is (assume that the pressure remains constant)
139279
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given, $\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{s}$, $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ $\because \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \mathrm{M}=\frac{3 \mathrm{RT}}{\mathrm{v}_{\mathrm{rms}}^{2}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}=4 \times 10^{-3} \mathrm{~kg}=4 \mathrm{~g}$ Therefore, the gas is He.
MP PET-2008
Kinetic Theory of Gases
139280
Pressure of an ideal gas is increased by keeping temperature constant. What is effect on kinetic energy of molecules?
1 Increase
2 Decrease
3 No change
4 Cannot be determined
Explanation:
C $\because$ Kinetic energy of gas (K.E) $=\frac{3 \mathrm{KT}}{2}$ where, $\mathrm{K} \rightarrow$ Boltzmann constant $\mathrm{T} \rightarrow \text { Temperature }$ $\mathrm{K} . \mathrm{E} \propto \mathrm{T}$ Kinetic energy of gaseous molecules, it only depends on the temperature but independent of pressure and Volume of the gas. According to the question if temperature of the gas is constant, so the kinetic energy of molecules will be also remain same.
JIPMER-2011
Kinetic Theory of Gases
139281
When temperature of an ideal gas in increased from $27^{\circ} \mathrm{C}$ to $227^{\circ} \mathrm{C}$, its rms speed is changed from $400 \mathrm{~m} / \mathrm{s}$ to $v_{\mathrm{s}}$. Then, $v_{\mathrm{s}}$ is
139275
The rms velocity of a gas at $T^{\circ} \mathrm{C}$ is double the value at $27^{\circ} \mathrm{C}$. The temperature $\mathrm{T}$ of the gas in ${ }^{0} \mathrm{C}$ is (assume that the pressure remains constant)
139279
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given, $\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{s}$, $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ $\because \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \mathrm{M}=\frac{3 \mathrm{RT}}{\mathrm{v}_{\mathrm{rms}}^{2}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}=4 \times 10^{-3} \mathrm{~kg}=4 \mathrm{~g}$ Therefore, the gas is He.
MP PET-2008
Kinetic Theory of Gases
139280
Pressure of an ideal gas is increased by keeping temperature constant. What is effect on kinetic energy of molecules?
1 Increase
2 Decrease
3 No change
4 Cannot be determined
Explanation:
C $\because$ Kinetic energy of gas (K.E) $=\frac{3 \mathrm{KT}}{2}$ where, $\mathrm{K} \rightarrow$ Boltzmann constant $\mathrm{T} \rightarrow \text { Temperature }$ $\mathrm{K} . \mathrm{E} \propto \mathrm{T}$ Kinetic energy of gaseous molecules, it only depends on the temperature but independent of pressure and Volume of the gas. According to the question if temperature of the gas is constant, so the kinetic energy of molecules will be also remain same.
JIPMER-2011
Kinetic Theory of Gases
139281
When temperature of an ideal gas in increased from $27^{\circ} \mathrm{C}$ to $227^{\circ} \mathrm{C}$, its rms speed is changed from $400 \mathrm{~m} / \mathrm{s}$ to $v_{\mathrm{s}}$. Then, $v_{\mathrm{s}}$ is
139275
The rms velocity of a gas at $T^{\circ} \mathrm{C}$ is double the value at $27^{\circ} \mathrm{C}$. The temperature $\mathrm{T}$ of the gas in ${ }^{0} \mathrm{C}$ is (assume that the pressure remains constant)
139279
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given, $\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{s}$, $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ $\because \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \mathrm{M}=\frac{3 \mathrm{RT}}{\mathrm{v}_{\mathrm{rms}}^{2}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}=4 \times 10^{-3} \mathrm{~kg}=4 \mathrm{~g}$ Therefore, the gas is He.
MP PET-2008
Kinetic Theory of Gases
139280
Pressure of an ideal gas is increased by keeping temperature constant. What is effect on kinetic energy of molecules?
1 Increase
2 Decrease
3 No change
4 Cannot be determined
Explanation:
C $\because$ Kinetic energy of gas (K.E) $=\frac{3 \mathrm{KT}}{2}$ where, $\mathrm{K} \rightarrow$ Boltzmann constant $\mathrm{T} \rightarrow \text { Temperature }$ $\mathrm{K} . \mathrm{E} \propto \mathrm{T}$ Kinetic energy of gaseous molecules, it only depends on the temperature but independent of pressure and Volume of the gas. According to the question if temperature of the gas is constant, so the kinetic energy of molecules will be also remain same.
JIPMER-2011
Kinetic Theory of Gases
139281
When temperature of an ideal gas in increased from $27^{\circ} \mathrm{C}$ to $227^{\circ} \mathrm{C}$, its rms speed is changed from $400 \mathrm{~m} / \mathrm{s}$ to $v_{\mathrm{s}}$. Then, $v_{\mathrm{s}}$ is
139275
The rms velocity of a gas at $T^{\circ} \mathrm{C}$ is double the value at $27^{\circ} \mathrm{C}$. The temperature $\mathrm{T}$ of the gas in ${ }^{0} \mathrm{C}$ is (assume that the pressure remains constant)
139279
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given, $\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{s}$, $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ $\because \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \mathrm{M}=\frac{3 \mathrm{RT}}{\mathrm{v}_{\mathrm{rms}}^{2}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}=4 \times 10^{-3} \mathrm{~kg}=4 \mathrm{~g}$ Therefore, the gas is He.
MP PET-2008
Kinetic Theory of Gases
139280
Pressure of an ideal gas is increased by keeping temperature constant. What is effect on kinetic energy of molecules?
1 Increase
2 Decrease
3 No change
4 Cannot be determined
Explanation:
C $\because$ Kinetic energy of gas (K.E) $=\frac{3 \mathrm{KT}}{2}$ where, $\mathrm{K} \rightarrow$ Boltzmann constant $\mathrm{T} \rightarrow \text { Temperature }$ $\mathrm{K} . \mathrm{E} \propto \mathrm{T}$ Kinetic energy of gaseous molecules, it only depends on the temperature but independent of pressure and Volume of the gas. According to the question if temperature of the gas is constant, so the kinetic energy of molecules will be also remain same.
JIPMER-2011
Kinetic Theory of Gases
139281
When temperature of an ideal gas in increased from $27^{\circ} \mathrm{C}$ to $227^{\circ} \mathrm{C}$, its rms speed is changed from $400 \mathrm{~m} / \mathrm{s}$ to $v_{\mathrm{s}}$. Then, $v_{\mathrm{s}}$ is
139275
The rms velocity of a gas at $T^{\circ} \mathrm{C}$ is double the value at $27^{\circ} \mathrm{C}$. The temperature $\mathrm{T}$ of the gas in ${ }^{0} \mathrm{C}$ is (assume that the pressure remains constant)
139279
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given, $\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{s}$, $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ $\because \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \mathrm{M}=\frac{3 \mathrm{RT}}{\mathrm{v}_{\mathrm{rms}}^{2}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}=4 \times 10^{-3} \mathrm{~kg}=4 \mathrm{~g}$ Therefore, the gas is He.
MP PET-2008
Kinetic Theory of Gases
139280
Pressure of an ideal gas is increased by keeping temperature constant. What is effect on kinetic energy of molecules?
1 Increase
2 Decrease
3 No change
4 Cannot be determined
Explanation:
C $\because$ Kinetic energy of gas (K.E) $=\frac{3 \mathrm{KT}}{2}$ where, $\mathrm{K} \rightarrow$ Boltzmann constant $\mathrm{T} \rightarrow \text { Temperature }$ $\mathrm{K} . \mathrm{E} \propto \mathrm{T}$ Kinetic energy of gaseous molecules, it only depends on the temperature but independent of pressure and Volume of the gas. According to the question if temperature of the gas is constant, so the kinetic energy of molecules will be also remain same.
JIPMER-2011
Kinetic Theory of Gases
139281
When temperature of an ideal gas in increased from $27^{\circ} \mathrm{C}$ to $227^{\circ} \mathrm{C}$, its rms speed is changed from $400 \mathrm{~m} / \mathrm{s}$ to $v_{\mathrm{s}}$. Then, $v_{\mathrm{s}}$ is