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An ideal gas follows aprocess described by $P V^{2}=C$ from $\left(P_{1}, V_{1}, T_{1}\right)$ to $\left(P_{2}, V_{2}, T_{2}\right)(C$ is a constant). Then
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The correct answer is:
if $\mathrm{V}_{2}^{1}>\mathrm{V}_{1}^{2}$ then $\mathrm{T}_{2}^{2} < \mathrm{T}_{1}^{1}$
$\begin{array}{l}
P V^{2}=C \\
P V . V=K \\
nR T . V=K \\
V T=K \\
V_{1} T_{1}=V_{2} T_{2} \\
V_{2}>V_{1} T < T \\
P^{2} 1 V_{1}^{2}=P 2 V_{2}^{2} \\
\frac{V_{2}^{2}}{V_{1}^{2}}=\frac{P_{1}}{P_{2}} \\
\text { P. } \frac{K^{2}}{T^{2}}=C \\
\frac{P}{T^{2}}=k \\
\frac{P_{1}}{T_{1}^{2}}=\frac{T_{1}^{2}}{T_{2}^{2}} \\
\frac{P_{1}}{P_{2}}=\frac{T_{1}^{2}}{T_{2}^{2}}
\end{array}$
P V^{2}=C \\
P V . V=K \\
nR T . V=K \\
V T=K \\
V_{1} T_{1}=V_{2} T_{2} \\
V_{2}>V_{1} T < T \\
P^{2} 1 V_{1}^{2}=P 2 V_{2}^{2} \\
\frac{V_{2}^{2}}{V_{1}^{2}}=\frac{P_{1}}{P_{2}} \\
\text { P. } \frac{K^{2}}{T^{2}}=C \\
\frac{P}{T^{2}}=k \\
\frac{P_{1}}{T_{1}^{2}}=\frac{T_{1}^{2}}{T_{2}^{2}} \\
\frac{P_{1}}{P_{2}}=\frac{T_{1}^{2}}{T_{2}^{2}}
\end{array}$
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