If gamma is the ratio of molar specific heat | vedclass

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Question

(A) Given that $C_P/C_V = \gamma$.We know that $C_P - C_V = R$,so $C_V = \frac{R}{\gamma - 1}$.The change in internal energy for any process is given

Vedclass · Accepted Answer

$\frac{PV}{(\gamma - 1)}$

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(A) Given that $C_P/C_V = \gamma$.We know that $C_P - C_V = R$,so $C_V = \frac{R}{\gamma - 1}$.The change in internal energy for any process is given by $\Delta U = n C_V \Delta T$.Since $n = 1$,$\Delta U = C_V \Delta T = \frac{R \Delta T}{\gamma - 1}$.From the ideal gas equation $PV = nRT$,at constant pressure $P$,$P \Delta V = nR \Delta T$. Since $n = 1$,$P \Delta V = R \Delta T$.Substituting this into the internal energy equation: $\Delta U = \frac{P \Delta V}{\gamma - 1}$.Given $\Delta V = 2V - V = V$,we get $\Delta U = \frac{PV}{\gamma - 1}$.

If $\gamma$ is the ratio of molar specific heat at constant pressure to molar specific heat at constant volume for a gas,find the change in internal energy of $1 \, mol$ of the gas when its volume changes from $V$ to $2V$ at constant pressure $P$.

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