Ranking: work in expansions

Problem

Four processes on one mole of ideal gas expanding from $V_1$ to $V_2 = 2V_1$:

• (A) isothermal at $300\ \text{K}$;
• (B) isobaric at $p_1 = 1\ \text{atm}$ starting from $T = 300\ \text{K}$;
• (C) isochoric followed by isobaric up to the same $V_2$;
• (D) adiabatic with $\gamma = 7/5$ starting from $T = 300\ \text{K}$.

Rank the work done by the gas from smallest to largest.

Solution

(A) Isothermal. The work is

$L_A = nRT\ln\frac{V_2}{V_1} = 8{,}31\cdot 300\cdot \ln 2 = 2493\cdot 0{,}693 \approx 1728\ \text{J}$

(B) Isobaric. At constant pressure, if $V$ doubles then $T$ also doubles ($T_2 = 600\ \text{K}$). The work is

$L_B = p_1\,\Delta V = nR\,\Delta T = nR\,T_1 = 8{,}31\cdot 300 \approx 2493\ \text{J}$

(D) Adiabatic. The final temperature is

$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300\cdot 0{,}5^{0{,}4} \approx 227\ \text{K}$

and, since $Q=0$, the work comes from the internal energy:

$L_D = n\,C_V(T_1 - T_2) = \frac{5}{2}\cdot 8{,}31\cdot(300-227) = 20{,}78\cdot 73 \approx 1517\ \text{J}$

less than the isothermal case, because the gas cools and the pressure drops quickly.

(C) Isochoric + isobaric. In the first stretch (isochoric) the volume does not change and $L=0$; then the final isobaric stretch pushes up to $V_2$ at a pressure $\ge p_1$ (because the isochoric stretch first increased $p$). The work is therefore comparable to or slightly greater than that of the isobaric process $B$.

Ranking.

$\ev{D < A < B \approx C}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes Skills: Ranking reasoning Objects: Ideal gas