Let's suppose a monoatomic gas of $n$ moles has been expanded from $V_1$ to $V_2$ volume keeping pressure constant. Now from the ideal gas law,we have $PV=nRT$. Taking differentials of both sides,we get $PdV=nRdT$. Here $PdV$ is our work done since the pressure is constant. Therefore,our work done should also be $nRdT$. But again,we are familiar with the energy of $n$ moles of gas at temperature $T$,which is $\frac{3}{2}nRT$. And we know,work done=change in kinetic energy. Therefore,our work done should be $\frac{3}{2}nRdT$. But earlier we argued,it should be $nRdT$. Now i am confused about which one is right. Please help me in identifying the wrong one and also please shed light on which one is wrong. Thanks in advance.
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$\begingroup$ You forgot to consider that heat has to be added to keep the pressure constant. $\endgroup$– Chet MillerJan 31 at 12:28
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$\begingroup$ Voting to reopen. This seems to be asking about a concept, namely the difference between work done and the change in internal energy. Just because the answer is basic doesn't make it a "homework-like question"! $\endgroup$– Michael SeifertJan 31 at 15:33
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$\begingroup$ Thank you for the support @MichaelSeifert. I really don't understand what fun they get in marking genuine concept based questions as homework questions without even understanding what i am trying to ask. Could you please shed some light on this topic as you said the difference between work done and internal energy change? It will be more than helpful. $\endgroup$– madnessJan 31 at 17:20
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$\begingroup$ @madness It's usually good to assume good faith rather than just assume it's people having fun shutting down questions. FWIW my vote to close was not "THIS IS HOMEWORK", but rather because it looked like the second and very overlooked condition of "check my work", as you have work written out, and you were asking for people to point out what is incorrect. Regardless, since others want it open I have supplied an answer (actually posted and deleted before the initial closure) $\endgroup$– BioPhysicistFeb 1 at 4:39
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Work done on a gas $W$ is not the only way to change the internal energy $U$ of an ideal gas; you are not accounting for the heat $Q$ needed for this process as well. The correct conservation of energy equation here (the First Law of Thermodynamics) is
$$\Delta U=W+Q$$
answered Jan 31 at 5:54