(A) For an adiabatic process, the relationship between temperature and volume is given by $T V^{\gamma - 1} = \text{constant}$.Since the gas is monoat

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(A) For an adiabatic process, the relationship between temperature and volume is given by $T V^{\gamma - 1} = \text{constant}$.Since the gas is monoatomic, the adiabatic index $\gamma = 5/3$.Therefore, $\gamma - 1 = 5/3 - 1 = 2/3$.The volume of the gas in a cylinder of cross-sectional area $A$ is $V = A \times L$.Thus, $T_1 (A L_1)^{2/3} = T_2 (A L_2)^{2/3}$.Rearranging the terms, we get $(T_2 / T_1) = (L_1 / L_2)^{2/3}$.

Class 11 Physics Thermodynamics Adiabatic Process

Medium

$A$ monoatomic ideal gas, initially at temperature $T_1$, is enclosed in a cylinder fitted with a massless, frictionless piston. By releasing the piston suddenly, the gas is allowed to expand adiabatically to a temperature $T_2$. If $L_1$ and $L_2$ are the lengths of the gas columns before and after expansion respectively, then $(T_2 / T_1)$ is given by

MHT CET 2025 All MHT CET

A
$(L_1 / L_2)^{2/3}$
B
$(L_2 / L_1)^{2/3}$
C
$(L_1 / L_2)$
D
$(L_2 / L_1)$

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The pressure and density of a given mass of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$ change adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d} = 32$, then $\frac{P^{\prime}}{P}$ is $(\gamma = \text{ratio of specific heats})$.

DifficultAP EAMCET 2004

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An ideal gas initially at $0^{\circ} C$ temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is $3/2$, the change in temperature due to the thermodynamics process is . . . . . . $K.$

MediumJEE MAIN 2025

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During an adiabatic process, the cube of the pressure is found to be inversely proportional to the fourth power of the volume. Then the ratio of specific heats is

MediumKCET 2007

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An ideal gas has a specific heat capacity at constant pressure of $\frac{11}{10} R$. If one mole of this ideal gas at $125^{\circ} C$ does $83 \,J$ of work adiabatically, then the final temperature of the gas would be (Universal gas constant, $R=8.3 \,J \,K^{-1} \,mol^{-1}$ ). (in $^{\circ} C$)

EasyAP EAMCET 2022

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$A$ sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is (given $\frac{C_{P}}{C_{V}}=\gamma=\frac{3}{2}$,$R=$ gas constant).

DifficultMHT CET 2023

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The pressure and density of a given mass of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$ change adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d} = 32$, then $\frac{P^{\prime}}{P}$ is $(\gamma = \text{ratio of specific heats})$.

An ideal gas initially at $0^{\circ} C$ temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is $3/2$, the change in temperature due to the thermodynamics process is . . . . . . $K.$

During an adiabatic process, the cube of the pressure is found to be inversely proportional to the fourth power of the volume. Then the ratio of specific heats is

An ideal gas has a specific heat capacity at constant pressure of $\frac{11}{10} R$. If one mole of this ideal gas at $125^{\circ} C$ does $83 \,J$ of work adiabatically, then the final temperature of the gas would be (Universal gas constant, $R=8.3 \,J \,K^{-1} \,mol^{-1}$ ). (in $^{\circ} C$)

$A$ sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is (given $\frac{C_{P}}{C_{V}}=\gamma=\frac{3}{2}$,$R=$ gas constant).

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