Question 647: Thermodynamics & KTG

Question

The change in the internal energy of the mass of gas, when the volume changes from V to 2 V at constant pressure P is $\left(\gamma=\frac{\mathrm{Cp}}{\mathrm{Cv}}\right)$

Accepted Answer

Change in internal energy is given by,
$$ \Delta U=n C_v \Delta T \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$
Given, $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ and $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$
$$ \begin{aligned} & 1+\frac{R}{C_V}=\gamma \ & C_V=\frac{R}{(\gamma-1)} \ 	herefore & \Delta U=n\left(\frac{R}{\gamma-1}ight) \Delta T\,\,\,\, \ldots[ From(i) ]\end{aligned} $$
$$ \begin{aligned} &	ext { Using } P \Delta V=n R \Delta T 	ext { under constant } P 	ext {, }\ &\Delta U=\frac{P \Delta V}{\gamma-1}=\frac{P(2 V-V)}{\gamma-1}=\frac{P V}{(\gamma-1)} \end{aligned} $$

Answer

$\frac{\mathrm{V}}{\mathrm{P}(\gamma-1)}$

Answer

$\frac{\mathrm{P}}{\mathrm{V}(\gamma-1)}$

Answer

$\frac{\mathrm{PV}}{\gamma+1}$

Answer

$\frac{\mathrm{PV}}{\gamma-1}$

Thermodynamics & KTG

The change in the internal energy of the mass of gas, when the volume changes from V to 2 V at constant pressure P is \left(\gamma=\frac{\mathrm{Cp}}{\mathrm{Cv}}\right)

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