Question

A perfect gas (0.1 mol) having $\bar{C}_v$ = 1.50R (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is (-) _______ J (nearest integer)

[Given: R = 0.082 L atm K$^{-1}$ mol$^{-1}$]

Answer 1

-304 J

Answer 2

1. Identify the processes:

   • 1 → 2: Isochoric (constant $V$); no work done.
   • 2 → 3: Isobaric (constant $P$); work done $w = P \Delta V$.
   • 3 → 4: Isochoric; no work done.

2. Calculate work for process 2 → 3:

   • At state 2: $V = 1000\, \text{cm}^3 = 1\, \text{L}$
   • At state 3: $V = 2000\, \text{cm}^3 = 2\, \text{L}$
   • Pressure $P = 3.00\, \text{atm}$
   • Work (in L·atm): $$w_{2\to3} = 3.00 \, \text{atm} \times (2\, \text{L} - 1\, \text{L}) = 3.00\, \text{L·atm}$$

3. Convert work into Joules:

   • Use conversion: $1\, \text{L·atm} = 101.325\, \text{J}$ $$w_{2\to3} = 3.00 \times 101.325 \, \text{J} \approx 303.975 \, \text{J}$$
   • The gas expands in process 2 → 3, so by convention the work done by the system is negative: $$w = -304 \, \text{J} \quad (\text{nearest integer})$$

Explanation (Minimal Core):

Only process 2 → 3 does work: $w = P\Delta V = 3\, \text{atm}\times1\, \text{L} = 3\, \text{L·atm} = 303.975\, \text{J}$, and with expansion, $w$ is negative, so $w \approx -304 \, \text{J}$.