139081
An ideal gas at $27^{\circ} \mathrm{C}$ is compressed adiabatically to $\frac{8}{27}$ of its original volume. The rise in temperature is $\left(\gamma=\frac{5}{3}\right)$
1 $475^{\circ} \mathrm{C}$
2 $402^{\circ} \mathrm{C}$
3 $275^{\circ} \mathrm{C}$
4 $375^{\circ} \mathrm{C}$
Explanation:
D Given, $\gamma=\frac{5}{3}$ $\mathrm{~T}_{1}=27^{\circ} \mathrm{C}$ $\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{8}{27}$ According to adiabatic process- $\mathrm{TV}^{\gamma-1}=\mathrm{C}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{27}{8}\right)^{\frac{5}{3}-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{27}{8}\right)^{\frac{2}{3}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{9}{4}$ $\mathrm{T}_{2}=\frac{9}{4} \times \mathrm{T}_{1}$ $\mathrm{T}_{2} =\frac{9}{4} \times 300 \quad\left[\mathrm{~T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\right]$ $\mathrm{T}_{2} =675$ Rise in temperature- $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=675-300=375 \mathrm{~K}=375^{\circ} \mathrm{C}$
AIPMT-1999
Kinetic Theory of Gases
139082
The pressure $p$ for a gas is plotted against its absolute temperatures $T$ for two different volumes $V_{1}$ and $V_{2}$ where $V_{1}>V_{2}$.If $p$ is plotted on $\mathrm{y}$-axis and $\mathrm{T}$ on $\mathrm{x}$-axis, then
1 The curve for $V_{1}$ has greater slope than that for $\mathrm{V}_{2}$
2 The curve for $\mathrm{V}_{2}$ has greater slope than that for $\mathrm{V}_{1}$
3 Both curves have same slope
4 The curves intersect at some point other than $\mathrm{T}=0$
Explanation:
B At constant volume, $\mathrm{P} \propto \mathrm{T}$ According to an ideal gas equation- $\because \quad \mathrm{PV}=\mathrm{nRT}$ $\mathrm{P}=\left(\frac{\mathrm{nR}}{\mathrm{V}}\right) \mathrm{T}$ Then, slope $(\mathrm{m})=\frac{\mathrm{nR}}{\mathrm{V}}$ Hence, curve for $\mathrm{V}_{2}$ has greater slope than that for $\mathrm{V}_{1}$.
AP EMCET(Medical)-2010
Kinetic Theory of Gases
139083
An ideal gas is initially at temperature $T$ and volume $V$. Its volume is increased by $\Delta V$, due to an increase in temperature $\Delta T$, pressure remaining constant. The physical quantity $\delta=\frac{\Delta V}{V \Delta T}$ varies with temperature as
1
2
3
4
Explanation:
C According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ At constant pressure, $\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$ $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{P}}$ Putting the value of $P$, we get- $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{nRT} / \mathrm{V}}$ $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{V}}{\mathrm{T}}$ $\text { or } \quad \frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}=\frac{1}{\mathrm{~T}}=\delta$ Hence, $\delta$ is inversely proportional to temperature (T). So, the $\delta-\mathrm{T}$ graph will be rectangular hyperbola.
AP EMCET(Medical)-2010
Kinetic Theory of Gases
139084
If the pressure of an ideal gas contained in a closed vessel is increased by $0.5 \%$ the increase in temperature is $2^{0} \mathrm{C}$. the initial temperature of the gas is:
1 $27^{0} \mathrm{C}$
2 $127^{\circ} \mathrm{C}$
3 $300^{\circ} \mathrm{C}$
4 $400^{\circ} \mathrm{C}$
Explanation:
B Given, $\mathrm{P}_{1}=\mathrm{P}, \mathrm{P}_{2}=\mathrm{P}+0.5 \% \mathrm{P}=1.005 \mathrm{P}$ $\mathrm{T}_{1}=\mathrm{T}, \mathrm{T}_{2}=\mathrm{T}+2$ According to Gay-Lussac law- $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ Putting these value, we get- $\frac{P}{1.005 \mathrm{P}}=\frac{\mathrm{T}}{\mathrm{T}+2}$ $\frac{1.005}{1}=\frac{\mathrm{T}+2}{\mathrm{~T}}$ $1.005=1+\frac{2}{\mathrm{~T}}$ $\frac{2}{\mathrm{~T}}=0.005$ $\mathrm{~T}=\frac{2}{0.005}=400 \mathrm{~K}$ $\mathrm{~T}=(400-273)=127^{\circ} \mathrm{C}$
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Kinetic Theory of Gases
139081
An ideal gas at $27^{\circ} \mathrm{C}$ is compressed adiabatically to $\frac{8}{27}$ of its original volume. The rise in temperature is $\left(\gamma=\frac{5}{3}\right)$
1 $475^{\circ} \mathrm{C}$
2 $402^{\circ} \mathrm{C}$
3 $275^{\circ} \mathrm{C}$
4 $375^{\circ} \mathrm{C}$
Explanation:
D Given, $\gamma=\frac{5}{3}$ $\mathrm{~T}_{1}=27^{\circ} \mathrm{C}$ $\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{8}{27}$ According to adiabatic process- $\mathrm{TV}^{\gamma-1}=\mathrm{C}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{27}{8}\right)^{\frac{5}{3}-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{27}{8}\right)^{\frac{2}{3}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{9}{4}$ $\mathrm{T}_{2}=\frac{9}{4} \times \mathrm{T}_{1}$ $\mathrm{T}_{2} =\frac{9}{4} \times 300 \quad\left[\mathrm{~T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\right]$ $\mathrm{T}_{2} =675$ Rise in temperature- $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=675-300=375 \mathrm{~K}=375^{\circ} \mathrm{C}$
AIPMT-1999
Kinetic Theory of Gases
139082
The pressure $p$ for a gas is plotted against its absolute temperatures $T$ for two different volumes $V_{1}$ and $V_{2}$ where $V_{1}>V_{2}$.If $p$ is plotted on $\mathrm{y}$-axis and $\mathrm{T}$ on $\mathrm{x}$-axis, then
1 The curve for $V_{1}$ has greater slope than that for $\mathrm{V}_{2}$
2 The curve for $\mathrm{V}_{2}$ has greater slope than that for $\mathrm{V}_{1}$
3 Both curves have same slope
4 The curves intersect at some point other than $\mathrm{T}=0$
Explanation:
B At constant volume, $\mathrm{P} \propto \mathrm{T}$ According to an ideal gas equation- $\because \quad \mathrm{PV}=\mathrm{nRT}$ $\mathrm{P}=\left(\frac{\mathrm{nR}}{\mathrm{V}}\right) \mathrm{T}$ Then, slope $(\mathrm{m})=\frac{\mathrm{nR}}{\mathrm{V}}$ Hence, curve for $\mathrm{V}_{2}$ has greater slope than that for $\mathrm{V}_{1}$.
AP EMCET(Medical)-2010
Kinetic Theory of Gases
139083
An ideal gas is initially at temperature $T$ and volume $V$. Its volume is increased by $\Delta V$, due to an increase in temperature $\Delta T$, pressure remaining constant. The physical quantity $\delta=\frac{\Delta V}{V \Delta T}$ varies with temperature as
1
2
3
4
Explanation:
C According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ At constant pressure, $\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$ $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{P}}$ Putting the value of $P$, we get- $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{nRT} / \mathrm{V}}$ $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{V}}{\mathrm{T}}$ $\text { or } \quad \frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}=\frac{1}{\mathrm{~T}}=\delta$ Hence, $\delta$ is inversely proportional to temperature (T). So, the $\delta-\mathrm{T}$ graph will be rectangular hyperbola.
AP EMCET(Medical)-2010
Kinetic Theory of Gases
139084
If the pressure of an ideal gas contained in a closed vessel is increased by $0.5 \%$ the increase in temperature is $2^{0} \mathrm{C}$. the initial temperature of the gas is:
1 $27^{0} \mathrm{C}$
2 $127^{\circ} \mathrm{C}$
3 $300^{\circ} \mathrm{C}$
4 $400^{\circ} \mathrm{C}$
Explanation:
B Given, $\mathrm{P}_{1}=\mathrm{P}, \mathrm{P}_{2}=\mathrm{P}+0.5 \% \mathrm{P}=1.005 \mathrm{P}$ $\mathrm{T}_{1}=\mathrm{T}, \mathrm{T}_{2}=\mathrm{T}+2$ According to Gay-Lussac law- $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ Putting these value, we get- $\frac{P}{1.005 \mathrm{P}}=\frac{\mathrm{T}}{\mathrm{T}+2}$ $\frac{1.005}{1}=\frac{\mathrm{T}+2}{\mathrm{~T}}$ $1.005=1+\frac{2}{\mathrm{~T}}$ $\frac{2}{\mathrm{~T}}=0.005$ $\mathrm{~T}=\frac{2}{0.005}=400 \mathrm{~K}$ $\mathrm{~T}=(400-273)=127^{\circ} \mathrm{C}$
139081
An ideal gas at $27^{\circ} \mathrm{C}$ is compressed adiabatically to $\frac{8}{27}$ of its original volume. The rise in temperature is $\left(\gamma=\frac{5}{3}\right)$
1 $475^{\circ} \mathrm{C}$
2 $402^{\circ} \mathrm{C}$
3 $275^{\circ} \mathrm{C}$
4 $375^{\circ} \mathrm{C}$
Explanation:
D Given, $\gamma=\frac{5}{3}$ $\mathrm{~T}_{1}=27^{\circ} \mathrm{C}$ $\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{8}{27}$ According to adiabatic process- $\mathrm{TV}^{\gamma-1}=\mathrm{C}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{27}{8}\right)^{\frac{5}{3}-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{27}{8}\right)^{\frac{2}{3}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{9}{4}$ $\mathrm{T}_{2}=\frac{9}{4} \times \mathrm{T}_{1}$ $\mathrm{T}_{2} =\frac{9}{4} \times 300 \quad\left[\mathrm{~T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\right]$ $\mathrm{T}_{2} =675$ Rise in temperature- $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=675-300=375 \mathrm{~K}=375^{\circ} \mathrm{C}$
AIPMT-1999
Kinetic Theory of Gases
139082
The pressure $p$ for a gas is plotted against its absolute temperatures $T$ for two different volumes $V_{1}$ and $V_{2}$ where $V_{1}>V_{2}$.If $p$ is plotted on $\mathrm{y}$-axis and $\mathrm{T}$ on $\mathrm{x}$-axis, then
1 The curve for $V_{1}$ has greater slope than that for $\mathrm{V}_{2}$
2 The curve for $\mathrm{V}_{2}$ has greater slope than that for $\mathrm{V}_{1}$
3 Both curves have same slope
4 The curves intersect at some point other than $\mathrm{T}=0$
Explanation:
B At constant volume, $\mathrm{P} \propto \mathrm{T}$ According to an ideal gas equation- $\because \quad \mathrm{PV}=\mathrm{nRT}$ $\mathrm{P}=\left(\frac{\mathrm{nR}}{\mathrm{V}}\right) \mathrm{T}$ Then, slope $(\mathrm{m})=\frac{\mathrm{nR}}{\mathrm{V}}$ Hence, curve for $\mathrm{V}_{2}$ has greater slope than that for $\mathrm{V}_{1}$.
AP EMCET(Medical)-2010
Kinetic Theory of Gases
139083
An ideal gas is initially at temperature $T$ and volume $V$. Its volume is increased by $\Delta V$, due to an increase in temperature $\Delta T$, pressure remaining constant. The physical quantity $\delta=\frac{\Delta V}{V \Delta T}$ varies with temperature as
1
2
3
4
Explanation:
C According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ At constant pressure, $\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$ $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{P}}$ Putting the value of $P$, we get- $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{nRT} / \mathrm{V}}$ $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{V}}{\mathrm{T}}$ $\text { or } \quad \frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}=\frac{1}{\mathrm{~T}}=\delta$ Hence, $\delta$ is inversely proportional to temperature (T). So, the $\delta-\mathrm{T}$ graph will be rectangular hyperbola.
AP EMCET(Medical)-2010
Kinetic Theory of Gases
139084
If the pressure of an ideal gas contained in a closed vessel is increased by $0.5 \%$ the increase in temperature is $2^{0} \mathrm{C}$. the initial temperature of the gas is:
1 $27^{0} \mathrm{C}$
2 $127^{\circ} \mathrm{C}$
3 $300^{\circ} \mathrm{C}$
4 $400^{\circ} \mathrm{C}$
Explanation:
B Given, $\mathrm{P}_{1}=\mathrm{P}, \mathrm{P}_{2}=\mathrm{P}+0.5 \% \mathrm{P}=1.005 \mathrm{P}$ $\mathrm{T}_{1}=\mathrm{T}, \mathrm{T}_{2}=\mathrm{T}+2$ According to Gay-Lussac law- $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ Putting these value, we get- $\frac{P}{1.005 \mathrm{P}}=\frac{\mathrm{T}}{\mathrm{T}+2}$ $\frac{1.005}{1}=\frac{\mathrm{T}+2}{\mathrm{~T}}$ $1.005=1+\frac{2}{\mathrm{~T}}$ $\frac{2}{\mathrm{~T}}=0.005$ $\mathrm{~T}=\frac{2}{0.005}=400 \mathrm{~K}$ $\mathrm{~T}=(400-273)=127^{\circ} \mathrm{C}$
139081
An ideal gas at $27^{\circ} \mathrm{C}$ is compressed adiabatically to $\frac{8}{27}$ of its original volume. The rise in temperature is $\left(\gamma=\frac{5}{3}\right)$
1 $475^{\circ} \mathrm{C}$
2 $402^{\circ} \mathrm{C}$
3 $275^{\circ} \mathrm{C}$
4 $375^{\circ} \mathrm{C}$
Explanation:
D Given, $\gamma=\frac{5}{3}$ $\mathrm{~T}_{1}=27^{\circ} \mathrm{C}$ $\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{8}{27}$ According to adiabatic process- $\mathrm{TV}^{\gamma-1}=\mathrm{C}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{27}{8}\right)^{\frac{5}{3}-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{27}{8}\right)^{\frac{2}{3}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{9}{4}$ $\mathrm{T}_{2}=\frac{9}{4} \times \mathrm{T}_{1}$ $\mathrm{T}_{2} =\frac{9}{4} \times 300 \quad\left[\mathrm{~T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\right]$ $\mathrm{T}_{2} =675$ Rise in temperature- $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=675-300=375 \mathrm{~K}=375^{\circ} \mathrm{C}$
AIPMT-1999
Kinetic Theory of Gases
139082
The pressure $p$ for a gas is plotted against its absolute temperatures $T$ for two different volumes $V_{1}$ and $V_{2}$ where $V_{1}>V_{2}$.If $p$ is plotted on $\mathrm{y}$-axis and $\mathrm{T}$ on $\mathrm{x}$-axis, then
1 The curve for $V_{1}$ has greater slope than that for $\mathrm{V}_{2}$
2 The curve for $\mathrm{V}_{2}$ has greater slope than that for $\mathrm{V}_{1}$
3 Both curves have same slope
4 The curves intersect at some point other than $\mathrm{T}=0$
Explanation:
B At constant volume, $\mathrm{P} \propto \mathrm{T}$ According to an ideal gas equation- $\because \quad \mathrm{PV}=\mathrm{nRT}$ $\mathrm{P}=\left(\frac{\mathrm{nR}}{\mathrm{V}}\right) \mathrm{T}$ Then, slope $(\mathrm{m})=\frac{\mathrm{nR}}{\mathrm{V}}$ Hence, curve for $\mathrm{V}_{2}$ has greater slope than that for $\mathrm{V}_{1}$.
AP EMCET(Medical)-2010
Kinetic Theory of Gases
139083
An ideal gas is initially at temperature $T$ and volume $V$. Its volume is increased by $\Delta V$, due to an increase in temperature $\Delta T$, pressure remaining constant. The physical quantity $\delta=\frac{\Delta V}{V \Delta T}$ varies with temperature as
1
2
3
4
Explanation:
C According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ At constant pressure, $\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$ $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{P}}$ Putting the value of $P$, we get- $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{nRT} / \mathrm{V}}$ $\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{V}}{\mathrm{T}}$ $\text { or } \quad \frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}=\frac{1}{\mathrm{~T}}=\delta$ Hence, $\delta$ is inversely proportional to temperature (T). So, the $\delta-\mathrm{T}$ graph will be rectangular hyperbola.
AP EMCET(Medical)-2010
Kinetic Theory of Gases
139084
If the pressure of an ideal gas contained in a closed vessel is increased by $0.5 \%$ the increase in temperature is $2^{0} \mathrm{C}$. the initial temperature of the gas is:
1 $27^{0} \mathrm{C}$
2 $127^{\circ} \mathrm{C}$
3 $300^{\circ} \mathrm{C}$
4 $400^{\circ} \mathrm{C}$
Explanation:
B Given, $\mathrm{P}_{1}=\mathrm{P}, \mathrm{P}_{2}=\mathrm{P}+0.5 \% \mathrm{P}=1.005 \mathrm{P}$ $\mathrm{T}_{1}=\mathrm{T}, \mathrm{T}_{2}=\mathrm{T}+2$ According to Gay-Lussac law- $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ Putting these value, we get- $\frac{P}{1.005 \mathrm{P}}=\frac{\mathrm{T}}{\mathrm{T}+2}$ $\frac{1.005}{1}=\frac{\mathrm{T}+2}{\mathrm{~T}}$ $1.005=1+\frac{2}{\mathrm{~T}}$ $\frac{2}{\mathrm{~T}}=0.005$ $\mathrm{~T}=\frac{2}{0.005}=400 \mathrm{~K}$ $\mathrm{~T}=(400-273)=127^{\circ} \mathrm{C}$