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A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 $\ce{dm^{3}}$ to a volume of 20 $\ce{dm^{3}}$ . It absorbs 800J of thermal energy from its surrounding. The change in internal energy is :-

I came across this question and my doubt is, for isothermal process, $\Delta U=0$ ( change in internal energy) for ideal gas is zero right? Then how can we calculate $\Delta U=0$ here?,assuming the gas to be ideal.

Is the error with the question or am I missing something?

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    $\begingroup$ $\Delta U=0$ for an isothermal process on an ideal gas. $\endgroup$ – Buck Thorn Aug 1 at 10:11
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    $\begingroup$ The process is irreversible, see the P-V graph, ext. P is at 1 but volume changes by 10 units implying a straight line. After this observation, proceed with dq=dw+du $\endgroup$ – Anindya Prithvi Aug 1 at 10:34
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    $\begingroup$ No, it is isothermal, just irreversible.... $\endgroup$ – Anindya Prithvi Aug 1 at 10:51
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    $\begingroup$ There are two types, one is reversible process and one is irreversible, it can be for adiabetic process as well, i.e. irreversible adiabetic process and reversible adiabetic process exist seperately. $\endgroup$ – Anindya Prithvi Aug 1 at 10:52
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    $\begingroup$ refer here chem.libretexts.org/Bookshelves/… $\endgroup$ – Anindya Prithvi Aug 1 at 10:55
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It must not be an ideal gas. This is irrespective or whether the process is reversible or irreversible. That is the only possible conclusion. Otherwise the amount of heat received would adjust until it was equal to the work done.

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