The volume of an ideal gas increases 8 times | vedclass

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(D) For a reversible adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.Let the init

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$ He $

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(D) For a reversible adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.Let the initial state be $(T_1, V_1)$ and the final state be $(T_2, V_2)$.Given: $V_2 = 8V_1$ and $T_2 = T_1/4$.Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Substituting the values: $T_1 V_1^{\gamma-1} = (T_1/4) (8V_1)^{\gamma-1}$.Dividing both sides by $T_1 V_1^{\gamma-1}$: $1 = (1/4) \cdot 8^{\gamma-1}$.$4 = 8^{\gamma-1}$.Expressing in base $2$: $2^2 = (2^3)^{\gamma-1} = 2^{3\gamma-3}$.Equating the exponents: $2 = 3\gamma - 3$.$3\gamma = 5$, which gives $\gamma = 5/3$.A gas with adiabatic index $\gamma = 5/3$ is a monoatomic gas.Among the given options, Helium (He) is a monoatomic gas. Therefore, the correct option is $D$.

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:

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