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(B) Given the equation of state: $P^2V = K$ (where $K$ is a constant).From the ideal gas law, $PV = nRT$, we have $P = \frac{nRT}{V}$.Substituting $P$

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$\sqrt{2} T$

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(B) Given the equation of state: $P^2V = K$ (where $K$ is a constant).From the ideal gas law, $PV = nRT$, we have $P = \frac{nRT}{V}$.Substituting $P$ into the given equation: $(\frac{nRT}{V})^2 V = K$.This simplifies to: $\frac{n^2 R^2 T^2}{V^2} \cdot V = K$, which gives $\frac{n^2 R^2 T^2}{V} = K$.Therefore, $T^2 \propto V$, or $T \propto \sqrt{V}$.Let the initial state be $(T_1, V_1) = (T, V)$ and the final state be $(T_2, V_2) = (T', 2V)$.Using the proportionality $\frac{T_1}{T_2} = \sqrt{\frac{V_1}{V_2}}$, we get $\frac{T}{T'} = \sqrt{\frac{V}{2V}} = \frac{1}{\sqrt{2}}$.Thus, $T' = \sqrt{2} T$.

During an experiment, an ideal gas obeys an additional equation of state $P^2V = \text{constant}$. The initial temperature and volume of the gas are $T$ and $V$ respectively. When it expands to volume $2V$, its temperature will be:

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