(B) For an adiabatic process,the work done $W$ on the gas is given by $W = \frac{nR\Delta T}{1 - \gamma}$.Here,$n = 1 \text{ kilo mole} = 1000 \text{

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(B) For an adiabatic process,the work done $W$ on the gas is given by $W = \frac{nR\Delta T}{1 - \gamma}$.Here,$n = 1 \text{ kilo mole} = 1000 \text{ moles}$,$R = 8.3 \ J \ mol^{-1} K^{-1}$,$\Delta T = 7 \ K$,and $W = -146 \ kJ = -146000 \ J$ (work is done on the gas).Substituting the values: $-146000 = \frac{1000 \times 8.3 \times 7}{1 - \gamma}$.$1 - \gamma = -\frac{58100}{146000} \approx -0.3979 \approx -0.4$.$1 - \gamma = -0.4 \Rightarrow \gamma = 1.4$.Since $\gamma = 1.4$ for a diatomic gas,the gas is diatomic.

Class 11 Physics Thermodynamics Adiabatic Process

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The work of $146 \ kJ$ is performed in order to compress one kilo mole of gas adiabatically,and in this process,the temperature of the gas increases by $7 ^\circ C$. The gas is $(R = 8.3 \ J \ mol^{-1} K^{-1})$.

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monoatomic
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diatomic
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triatomic
D
a mixture of monoatomic and diatomic

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$A$ monatomic gas at a pressure of $100 \text{ kPa}$ expands adiabatically such that its final volume becomes $8$ times its initial volume. If the work done during the process is $180 \text{ J}$, then the initial volume of the gas is (in $\text{ cm}^3$)

MediumAP EAMCET 2025

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In an adiabatic change,the pressure $P$ and temperature $T$ of a diatomic gas are related by the relation $P \propto T^C$,where $C$ is equal to:

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The volume of an ideal gas increases $8$ times and the temperature becomes $1/4$ of the initial temperature during a reversible adiabatic change. If there is no exchange of heat in this process $( \Delta Q=0 )$, identify the gas from the following options:

DifficultJEE MAIN 2026

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The $P-V$ diagram for two adiabatic processes is given. Curves $1$ and $2$ correspond to:

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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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The work of $146 \ kJ$ is performed in order to compress one kilo mole of gas adiabatically,and in this process,the temperature of the gas increases by $7 ^\circ C$. The gas is $(R = 8.3 \ J \ mol^{-1} K^{-1})$.

Similar Questions

$A$ monatomic gas at a pressure of $100 \text{ kPa}$ expands adiabatically such that its final volume becomes $8$ times its initial volume. If the work done during the process is $180 \text{ J}$, then the initial volume of the gas is (in $\text{ cm}^3$)

In an adiabatic change,the pressure $P$ and temperature $T$ of a diatomic gas are related by the relation $P \propto T^C$,where $C$ is equal to:

The volume of an ideal gas increases $8$ times and the temperature becomes $1/4$ of the initial temperature during a reversible adiabatic change. If there is no exchange of heat in this process $( \Delta Q=0 )$, identify the gas from the following options:

The $P-V$ diagram for two adiabatic processes is given. Curves $1$ and $2$ correspond to:

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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