Answer to Question #164389 in Molecular Physics | Thermodynamics for anna francisco

Question

10 mol of N₂ gas expands at constant temperature of 300 K from 2 L at 8 atm to 6 L at 4 atm. How much work is performed by the gas?

Answer 1

Since temperature is constant, process is isothermal. Work done in isothermal process is given by, $W = 2.303nRTlog(V_2/V_1)$

Putting value, $W=2.303\times 10\times 8.3\times 300log(6/2)=27.36KJ$