# Change in internal energy for isothermal process

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?

• $\Delta U=0$ for an isothermal process on an ideal gas. – Buck Thorn Aug 1 '20 at 10:11
• 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 – user96208 Aug 1 '20 at 10:34
• No, it is isothermal, just irreversible.... – user96208 Aug 1 '20 at 10:51
• 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. – user96208 Aug 1 '20 at 10:52
• refer here chem.libretexts.org/Bookshelves/… – user96208 Aug 1 '20 at 10:55