Thermodynamic Processes

Question

Two moles of an ideal gas undergo a reversible isothermal expansion at 27 °C from a volume of 10 L to 100 L. The work done by the gas is approximately: (R = 8.314 J K⁻¹ mol⁻¹)

RankGuru · Accepted Answer

1. **Formula for Isothermal Reversible Work**: $w_{	ext{by gas}} = nRT \ln\frac{V_2}{V_1}$ (work done *by* the gas is positive for expansion).
2. **Convert Temperature**: $T = 27 + 273 = 300$ K.
3. **Substitute**: $w = 2 	imes 8.314 	imes 300 	imes \ln\frac{100}{10}$
   $= 2 	imes 8.314 	imes 300 	imes \ln 10$
   $= 2 	imes 8.314 	imes 300 	imes 2.303$
   $= 2 	imes 8.314 	imes 300 	imes 2.303 \approx 11488$ J $\approx 11.49$ kJ.
4. **Sign**: Positive, since the gas expands and does work on the surroundings.
5. **For an Isothermal Process**: ΔU = 0 for an ideal gas, so q = w = +11.49 kJ — the gas absorbs this much heat from the surroundings to maintain constant temperature during expansion.

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