A perfect gas ( 0.1 mol ) having C̅ᵥ=1.50 R (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 ]

Solution

Chemistry ma’am · 4 steps

Only the flat top of the box does work: vertical lines on a P–V graph give zero work.

1. Read the path. Step 1→2 and step 3→4 are vertical lines, so the volume stays constant. With Δ V = 0, no work is done in either step.
2. Isobaric step. Step 2→3 is at a constant 3 atm. The volume goes from 1000 cm³ = 1 L to 2000 cm³ = 2 L, so the gas expands and does work on the surroundings.

   w = -Pₑₓₜ Δ V = -(3 atm)(2 - 1) L = -3 L atm

3. Convert to joules. Using 1 L atm = 101.325 J, w = -3 × 101.325 = -303.975 J. The value of C̅ᵥ is not needed, because it only matters for Δ U and q.
4. So. The total work from point 1 to point 4 is w ≈ -304 J, so the blank is 304.

The common slip: 3. Stopping at 3 L atm and reporting 3 forgets the conversion to joules. Multiply by 101.3 J per L atm.

The idea to keep. On a P–V diagram, work is the area under the path. Vertical (isochoric) lines add nothing, so find the horizontal parts and use w = -PΔ V. Convert L atm to joules at the end.

Answer from the NTA final answer key. Solution by PrepWiser.