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(A) This is a case of free expansion of an ideal gas into a vacuum.In free expansion,the gas does no work because it expands against zero external pre

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(A) This is a case of free expansion of an ideal gas into a vacuum.In free expansion,the gas does no work because it expands against zero external pressure,so $\Delta W = 0$.Since the system is thermally isolated from its surroundings,there is no heat exchange,so $\Delta Q = 0$.According to the first law of thermodynamics,$\Delta U = \Delta Q - \Delta W$.Substituting the values,$\Delta U = 0 - 0 = 0$.For an ideal gas,the internal energy $U$ is a function of temperature only $(U \propto T)$.Since $\Delta U = 0$,the change in temperature $\Delta T$ must also be $0$.Therefore,the final temperature remains the same as the initial temperature,which is $300 \ K$.

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