1.0 mol of monoatomic ideal gas is expanded from state 1 to state 2 as shown in the figure. The magnitude of the work done for the expansion of gas from state 1 to state 2 at 300 K is ____________ J. (Nearest integer)
(Given : R = 8.3 J K⁻¹ mol⁻¹, ln10 = 2.3, log2 = 0.30)

Solution

Chemistry ma’am · 4 steps

The curve is an isotherm at 300 K, so use the pressure ratio inside the log and stop there.

1. Read the process. The gas goes from 6.0 bar to 3.0 bar along a smooth curve at a fixed 300 K. That is a reversible isothermal expansion of an ideal gas.
2. Use the pressure form. At constant temperature PV is constant, so V₂/V₁ = P₁/P₂. The work done therefore needs only the pressure ratio.

   |W| = nRT ln P₁/P₂ = 1 × 8.3 × 300 × ln 6.0/3.0

3. Put in numbers. ln 2 = 2.3 × log 2 = 2.3 × 0.30 = 0.69. Also nRT = 8.3 × 300 = 2490 J, so |W| = 2490 × 0.69 = 1718.1 J.
4. So. The magnitude of the work done is about 1718 J.

The common slip: 747. Using log 2 = 0.30 directly instead of ln 2 = 0.69 gives 2490 × 0.30 = 747, not 1718. Multiply by 2.3 to convert log to ln.

The idea to keep. For a reversible isothermal change of an ideal gas, |W| = nRT ln V₂/V₁ = nRT ln P₁/P₂. Use whichever ratio the graph gives you, and turn log into ln with the factor 2.3.

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