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An ideal gas follows a process described by the equation $\mathrm{PV}^2=\mathrm{C}$ from the initial $\left(\mathrm{P}_1, \mathrm{~V}_1, \mathrm{~T}_1\right)$ to final ( $\left.\mathrm{P}_2, \mathrm{~V}_2, \mathrm{~T}_2\right)$ thermodynamic states, where $\mathrm{C}$ is a constant. Then:
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Verified Answer
The correct answer is:
If $\mathrm{V}_2>\mathrm{V}_1$ then $\mathrm{T}_2 < \mathrm{T}_1$
As per question,
$\mathrm{PV}^2=\mathrm{C}$
And by ideal gas equation, we have;
$\mathrm{PV}=n \mathrm{RT}$
Hence, $\left(\frac{n \mathrm{RT}}{\mathrm{V}}\right) \mathrm{V}^2=\mathrm{C}$
Hence, TV $=$ constant
$\mathrm{T} \propto \frac{1}{\mathrm{~V}} \therefore \mathrm{V}_2 > \mathrm{V}_1 \text { and } \mathrm{T}_1 > \mathrm{T}_2$
$\mathrm{PV}^2=\mathrm{C}$
And by ideal gas equation, we have;
$\mathrm{PV}=n \mathrm{RT}$
Hence, $\left(\frac{n \mathrm{RT}}{\mathrm{V}}\right) \mathrm{V}^2=\mathrm{C}$
Hence, TV $=$ constant
$\mathrm{T} \propto \frac{1}{\mathrm{~V}} \therefore \mathrm{V}_2 > \mathrm{V}_1 \text { and } \mathrm{T}_1 > \mathrm{T}_2$
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