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A gas expands isothermally against a constant external pressure of $1\ \mathrm{atm}$ from a volume of $10\ \mathrm{dm^3}$ to a volume of $20\ \mathrm{dm^3}$. It absorbs $800\ \mathrm J$ 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 ♦
    Commented Aug 1, 2020 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 $\endgroup$
    – user96208
    Commented Aug 1, 2020 at 10:34
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    $\begingroup$ No, it is isothermal, just irreversible.... $\endgroup$
    – user96208
    Commented Aug 1, 2020 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. $\endgroup$
    – user96208
    Commented Aug 1, 2020 at 10:52
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    $\begingroup$ refer here chem.libretexts.org/Bookshelves/… $\endgroup$
    – user96208
    Commented Aug 1, 2020 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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