Pressure over 1000 ml of a liquid is gradually increased from 1 bar to 1001 bar under adiabatic cond — Thermodynamics and Thermochemistry Chemistry Question
Question
Pressure over 1000 ml of a liquid is gradually increased from 1 bar to 1001 bar under adiabatic conditions. If the final volume of the liquid is 990 ml and there is linear variation of volume with pressure, the value of $\Delta U$ of the process is (Answer as ‘abcd’, where a = 1, if $\Delta U$ is +ve and a = 2, if $\Delta U$ is – ve, and ‘bcd’ is the magnitude of $\Delta U$, in J)
💡 Solution & Explanation
Adiabatic ($q = 0$). Gradual increase implies reversible work: $W = -\int P dV$. Linear variation implies $W = -\frac{1}{2}(P_1 + P_2)(V_2 - V_1) = -\frac{1}{2}(1 + 1001) \times 10^5 \times (990 - 1000) \times 10^{-6} = -501 \times 10^5 \times (-10 \times 10^{-6}) = 501\text{ J}$. $\Delta U = W = 501\text{ J}$. Positive means $a = 1$. Magnitude is 501, so 'abcd' = 1501.