(B) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.Given the condition

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(B) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.Given the condition $T \propto \frac{1}{\sqrt{V}}$,we can write this as $T \propto V^{-1/2}$,which implies $TV^{1/2} = \text{constant}$.Comparing the two expressions $TV^{\gamma-1} = \text{constant}$ and $TV^{1/2} = \text{constant}$,we get $\gamma - 1 = \frac{1}{2}$.Solving for $\gamma$,we find $\gamma = 1 + 0.5 = 1.5$.

Class 11 Physics Thermodynamics Adiabatic Process

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One mole of an ideal gas expands adiabatically at constant pressure such that its temperature $T \propto \frac{1}{\sqrt{V}}$. The value of $\gamma$ for the gas is $(\gamma = \frac{C_p}{C_v}, V = \text{Volume of the gas})$

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A
$1.8$
B
$1.5$
C
$1.3$
D
$1.4$

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Thermodynamics

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Adiabatic Process

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$A$ monoatomic ideal gas is compressed adiabatically to $\left(\frac{1}{27}\right)$ of its initial volume. If the initial temperature of the gas is $T \ K$ and the final temperature is $xT \ K$, the value of $x$ is:

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$A$ gas at $NTP$ is suddenly compressed to one-fourth of its original volume. If $\gamma$ is supposed to be $\frac{3}{2}$,then the final pressure is........ atmosphere.

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At $27^\circ C$,a gas is suddenly compressed such that its pressure becomes $1/8$th of the original pressure. The temperature of the gas will be $(\gamma = 5/3)$.

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In case of an adiabatic process,the correct relation in terms of pressure $p$ and density $\rho$ of a gas is:

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During an adiabatic expansion of $2\, moles$ of a gas,the change in internal energy was found to be $-50\, J$. The work done during the process is ...... $J$.

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One mole of an ideal gas expands adiabatically at constant pressure such that its temperature $T \propto \frac{1}{\sqrt{V}}$. The value of $\gamma$ for the gas is $(\gamma = \frac{C_p}{C_v}, V = \text{Volume of the gas})$

Similar Questions

$A$ monoatomic ideal gas is compressed adiabatically to $\left(\frac{1}{27}\right)$ of its initial volume. If the initial temperature of the gas is $T \ K$ and the final temperature is $xT \ K$, the value of $x$ is:

$A$ gas at $NTP$ is suddenly compressed to one-fourth of its original volume. If $\gamma$ is supposed to be $\frac{3}{2}$,then the final pressure is........ atmosphere.

At $27^\circ C$,a gas is suddenly compressed such that its pressure becomes $1/8$th of the original pressure. The temperature of the gas will be $(\gamma = 5/3)$.

In case of an adiabatic process,the correct relation in terms of pressure $p$ and density $\rho$ of a gas is:

During an adiabatic expansion of $2\, moles$ of a gas,the change in internal energy was found to be $-50\, J$. The work done during the process is ...... $J$.

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