139075
Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} \mathrm{C}$, if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K})$
139077
A vessel contains $32 \mathrm{gm}$ of $\mathrm{O}_{2}$ at a temperature $T$. The pressure of the gas is $P$. An identical vessel containing $4 \mathrm{gm}$ of $\mathrm{H}_{2}$ at a temperature $2 \mathrm{~T}$ has a pressure of
1 $8 \mathrm{P}$
2 $4 \mathrm{P}$
3 $\mathrm{P}$
4 $\frac{P}{8}$
Explanation:
B Given that, an identical container so volume is constant from ideal gas equation $\mathrm{PV}=\mathrm{nRT} \quad\{\because \quad \mathrm{V} \text { and } \mathrm{R} \text { are Constant }\}$ $\mathrm{P} \propto \mathrm{nT}$ Then, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{n}_{1} \mathrm{~T}_{1}}{\mathrm{n}_{2} \mathrm{~T}_{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\frac{32}{32}(\mathrm{~T})}{\frac{4}{2}(2 \mathrm{~T})}=\frac{1}{4} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}\right]$ $\mathrm{P}_{2}=4 \mathrm{P}$
AMU-2009
Kinetic Theory of Gases
139078
A cylinder contains 12 litres of oxygen at $20^{\circ} \mathrm{C}$ and $15 \mathrm{~atm}$ pressure. The temperature of the gas is raised to $35^{\circ} \mathrm{C}$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?
139079
A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be
1 $4 \mathrm{P}_{\mathrm{o}}$
2 $2 \mathrm{P}_{\mathrm{o}}$
3 $\mathrm{P}_{\mathrm{o}}$
4 $\frac{P_{o}}{2}$
Explanation:
B Given, $\mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}, \mathrm{~V}_{2}=2 \mathrm{~V}_{1}$ We know that $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{V}_{\mathrm{rms}}^{2}$ $\mathrm{P} \propto m \mathrm{~V}_{\mathrm{rms}}^{2}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}} \times\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{2}$ $=\frac{\mathrm{m}_{1} / 2}{\mathrm{~m}_{1}}\left(\frac{2 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{2} \quad\left[\because \mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}\right]$ $\mathrm{P}_{2}=2 \mathrm{P}_{1}$ $\mathrm{P}_{2}=2 \mathrm{P}_{\mathrm{o}} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}_{\mathrm{o}}\right]$
2004]
Kinetic Theory of Gases
139080
If a given mass of a gas occupies a volume $100 \mathrm{~cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} \mathrm{C}$, what will be its volume at 4 atmospheric pressure the temperature being the same
139075
Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} \mathrm{C}$, if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K})$
139077
A vessel contains $32 \mathrm{gm}$ of $\mathrm{O}_{2}$ at a temperature $T$. The pressure of the gas is $P$. An identical vessel containing $4 \mathrm{gm}$ of $\mathrm{H}_{2}$ at a temperature $2 \mathrm{~T}$ has a pressure of
1 $8 \mathrm{P}$
2 $4 \mathrm{P}$
3 $\mathrm{P}$
4 $\frac{P}{8}$
Explanation:
B Given that, an identical container so volume is constant from ideal gas equation $\mathrm{PV}=\mathrm{nRT} \quad\{\because \quad \mathrm{V} \text { and } \mathrm{R} \text { are Constant }\}$ $\mathrm{P} \propto \mathrm{nT}$ Then, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{n}_{1} \mathrm{~T}_{1}}{\mathrm{n}_{2} \mathrm{~T}_{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\frac{32}{32}(\mathrm{~T})}{\frac{4}{2}(2 \mathrm{~T})}=\frac{1}{4} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}\right]$ $\mathrm{P}_{2}=4 \mathrm{P}$
AMU-2009
Kinetic Theory of Gases
139078
A cylinder contains 12 litres of oxygen at $20^{\circ} \mathrm{C}$ and $15 \mathrm{~atm}$ pressure. The temperature of the gas is raised to $35^{\circ} \mathrm{C}$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?
139079
A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be
1 $4 \mathrm{P}_{\mathrm{o}}$
2 $2 \mathrm{P}_{\mathrm{o}}$
3 $\mathrm{P}_{\mathrm{o}}$
4 $\frac{P_{o}}{2}$
Explanation:
B Given, $\mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}, \mathrm{~V}_{2}=2 \mathrm{~V}_{1}$ We know that $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{V}_{\mathrm{rms}}^{2}$ $\mathrm{P} \propto m \mathrm{~V}_{\mathrm{rms}}^{2}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}} \times\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{2}$ $=\frac{\mathrm{m}_{1} / 2}{\mathrm{~m}_{1}}\left(\frac{2 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{2} \quad\left[\because \mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}\right]$ $\mathrm{P}_{2}=2 \mathrm{P}_{1}$ $\mathrm{P}_{2}=2 \mathrm{P}_{\mathrm{o}} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}_{\mathrm{o}}\right]$
2004]
Kinetic Theory of Gases
139080
If a given mass of a gas occupies a volume $100 \mathrm{~cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} \mathrm{C}$, what will be its volume at 4 atmospheric pressure the temperature being the same
139075
Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} \mathrm{C}$, if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K})$
139077
A vessel contains $32 \mathrm{gm}$ of $\mathrm{O}_{2}$ at a temperature $T$. The pressure of the gas is $P$. An identical vessel containing $4 \mathrm{gm}$ of $\mathrm{H}_{2}$ at a temperature $2 \mathrm{~T}$ has a pressure of
1 $8 \mathrm{P}$
2 $4 \mathrm{P}$
3 $\mathrm{P}$
4 $\frac{P}{8}$
Explanation:
B Given that, an identical container so volume is constant from ideal gas equation $\mathrm{PV}=\mathrm{nRT} \quad\{\because \quad \mathrm{V} \text { and } \mathrm{R} \text { are Constant }\}$ $\mathrm{P} \propto \mathrm{nT}$ Then, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{n}_{1} \mathrm{~T}_{1}}{\mathrm{n}_{2} \mathrm{~T}_{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\frac{32}{32}(\mathrm{~T})}{\frac{4}{2}(2 \mathrm{~T})}=\frac{1}{4} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}\right]$ $\mathrm{P}_{2}=4 \mathrm{P}$
AMU-2009
Kinetic Theory of Gases
139078
A cylinder contains 12 litres of oxygen at $20^{\circ} \mathrm{C}$ and $15 \mathrm{~atm}$ pressure. The temperature of the gas is raised to $35^{\circ} \mathrm{C}$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?
139079
A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be
1 $4 \mathrm{P}_{\mathrm{o}}$
2 $2 \mathrm{P}_{\mathrm{o}}$
3 $\mathrm{P}_{\mathrm{o}}$
4 $\frac{P_{o}}{2}$
Explanation:
B Given, $\mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}, \mathrm{~V}_{2}=2 \mathrm{~V}_{1}$ We know that $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{V}_{\mathrm{rms}}^{2}$ $\mathrm{P} \propto m \mathrm{~V}_{\mathrm{rms}}^{2}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}} \times\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{2}$ $=\frac{\mathrm{m}_{1} / 2}{\mathrm{~m}_{1}}\left(\frac{2 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{2} \quad\left[\because \mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}\right]$ $\mathrm{P}_{2}=2 \mathrm{P}_{1}$ $\mathrm{P}_{2}=2 \mathrm{P}_{\mathrm{o}} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}_{\mathrm{o}}\right]$
2004]
Kinetic Theory of Gases
139080
If a given mass of a gas occupies a volume $100 \mathrm{~cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} \mathrm{C}$, what will be its volume at 4 atmospheric pressure the temperature being the same
139075
Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} \mathrm{C}$, if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K})$
139077
A vessel contains $32 \mathrm{gm}$ of $\mathrm{O}_{2}$ at a temperature $T$. The pressure of the gas is $P$. An identical vessel containing $4 \mathrm{gm}$ of $\mathrm{H}_{2}$ at a temperature $2 \mathrm{~T}$ has a pressure of
1 $8 \mathrm{P}$
2 $4 \mathrm{P}$
3 $\mathrm{P}$
4 $\frac{P}{8}$
Explanation:
B Given that, an identical container so volume is constant from ideal gas equation $\mathrm{PV}=\mathrm{nRT} \quad\{\because \quad \mathrm{V} \text { and } \mathrm{R} \text { are Constant }\}$ $\mathrm{P} \propto \mathrm{nT}$ Then, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{n}_{1} \mathrm{~T}_{1}}{\mathrm{n}_{2} \mathrm{~T}_{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\frac{32}{32}(\mathrm{~T})}{\frac{4}{2}(2 \mathrm{~T})}=\frac{1}{4} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}\right]$ $\mathrm{P}_{2}=4 \mathrm{P}$
AMU-2009
Kinetic Theory of Gases
139078
A cylinder contains 12 litres of oxygen at $20^{\circ} \mathrm{C}$ and $15 \mathrm{~atm}$ pressure. The temperature of the gas is raised to $35^{\circ} \mathrm{C}$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?
139079
A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be
1 $4 \mathrm{P}_{\mathrm{o}}$
2 $2 \mathrm{P}_{\mathrm{o}}$
3 $\mathrm{P}_{\mathrm{o}}$
4 $\frac{P_{o}}{2}$
Explanation:
B Given, $\mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}, \mathrm{~V}_{2}=2 \mathrm{~V}_{1}$ We know that $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{V}_{\mathrm{rms}}^{2}$ $\mathrm{P} \propto m \mathrm{~V}_{\mathrm{rms}}^{2}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}} \times\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{2}$ $=\frac{\mathrm{m}_{1} / 2}{\mathrm{~m}_{1}}\left(\frac{2 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{2} \quad\left[\because \mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}\right]$ $\mathrm{P}_{2}=2 \mathrm{P}_{1}$ $\mathrm{P}_{2}=2 \mathrm{P}_{\mathrm{o}} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}_{\mathrm{o}}\right]$
2004]
Kinetic Theory of Gases
139080
If a given mass of a gas occupies a volume $100 \mathrm{~cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} \mathrm{C}$, what will be its volume at 4 atmospheric pressure the temperature being the same
139075
Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} \mathrm{C}$, if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K})$
139077
A vessel contains $32 \mathrm{gm}$ of $\mathrm{O}_{2}$ at a temperature $T$. The pressure of the gas is $P$. An identical vessel containing $4 \mathrm{gm}$ of $\mathrm{H}_{2}$ at a temperature $2 \mathrm{~T}$ has a pressure of
1 $8 \mathrm{P}$
2 $4 \mathrm{P}$
3 $\mathrm{P}$
4 $\frac{P}{8}$
Explanation:
B Given that, an identical container so volume is constant from ideal gas equation $\mathrm{PV}=\mathrm{nRT} \quad\{\because \quad \mathrm{V} \text { and } \mathrm{R} \text { are Constant }\}$ $\mathrm{P} \propto \mathrm{nT}$ Then, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{n}_{1} \mathrm{~T}_{1}}{\mathrm{n}_{2} \mathrm{~T}_{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\frac{32}{32}(\mathrm{~T})}{\frac{4}{2}(2 \mathrm{~T})}=\frac{1}{4} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}\right]$ $\mathrm{P}_{2}=4 \mathrm{P}$
AMU-2009
Kinetic Theory of Gases
139078
A cylinder contains 12 litres of oxygen at $20^{\circ} \mathrm{C}$ and $15 \mathrm{~atm}$ pressure. The temperature of the gas is raised to $35^{\circ} \mathrm{C}$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?
139079
A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be
1 $4 \mathrm{P}_{\mathrm{o}}$
2 $2 \mathrm{P}_{\mathrm{o}}$
3 $\mathrm{P}_{\mathrm{o}}$
4 $\frac{P_{o}}{2}$
Explanation:
B Given, $\mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}, \mathrm{~V}_{2}=2 \mathrm{~V}_{1}$ We know that $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{V}_{\mathrm{rms}}^{2}$ $\mathrm{P} \propto m \mathrm{~V}_{\mathrm{rms}}^{2}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}} \times\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{2}$ $=\frac{\mathrm{m}_{1} / 2}{\mathrm{~m}_{1}}\left(\frac{2 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{2} \quad\left[\because \mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{2}\right]$ $\mathrm{P}_{2}=2 \mathrm{P}_{1}$ $\mathrm{P}_{2}=2 \mathrm{P}_{\mathrm{o}} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}_{\mathrm{o}}\right]$
2004]
Kinetic Theory of Gases
139080
If a given mass of a gas occupies a volume $100 \mathrm{~cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} \mathrm{C}$, what will be its volume at 4 atmospheric pressure the temperature being the same
