• # question_answer A gas expands isothermally against a constant external pressure of 1 atm from a volume of $\text{10 d}{{\text{m}}^{\text{3}}}$to a volume of $\text{20 d}{{\text{m}}^{\text{3}}}\text{.}$ It absorbs 300 J of thermal energy from its surroundings. The $\Delta U$is A)  -312 J                        B)  -213 JC)  -321 J                        D)  -123 J

$w=--{{V}_{1}}\int_{{}}^{{}}{{{V}_{2}}.PdV=-P({{V}_{2}}-{{V}_{1}})}$ $W=-(1atm)(20-10)=-10\,d{{m}^{3}}\,\text{atm}\text{.}$ $=-10\,d{{m}^{3}}\times \frac{8.314\,J{{K}^{-1}}\,mo{{l}^{-1}}}{0.08206\,d{{m}^{3}}{{K}^{-1}}mo{{l}^{-1}}}$ $=-1013\,J.$ According to the first law of thermodynamics $\Delta U=q+w$ $=800+(-1013=-213\,J)$ Hence, the correct option is [b].