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(C) In a rigid container,the volume of the gas remains constant,which means $\Delta V = 0$. Since work done $W = -P_{ext} \Delta V$,it follows that $W

Vedclass · Accepted Answer

It is isobaric process

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(C) In a rigid container,the volume of the gas remains constant,which means $\Delta V = 0$. Since work done $W = -P_{ext} \Delta V$,it follows that $W = 0$. This type of process is known as an isochoric process. According to the first law of thermodynamics,$\Delta U = q + W$. Since $W = 0$,$\Delta U = q$,meaning the absorbed heat increases the internal energy of the system. An isobaric process is one where pressure remains constant,which is not the case here. Therefore,the statement that it is an isobaric process is incorrect.

An ideal gas is heated in a rigid container so its temperature rises. Which is incorrect for this process?

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