Posted by **Sandhya** on Thursday, December 4, 2008 at 6:05am.

5 mol of an ideal gas is kept at 394◦C in an expansion from 1 L to 5 L .How much work is done by the gas? Answer in units of J.Given: R = 8.31 J/K/mol ,

- Thermodynamics -
**drwls**, Thursday, December 4, 2008 at 7:01am
Hmm. An isothermal expansion. The pressure must fall by a factor of 5 during this process. The initial pressure is

P1 = n R T = (5 moles/liter)*0.08206 Liter atm/mole K)* 667 K = 274 atm = 2.77*10^7 N/m^2

P2, the final pressure, is 1/5 of P1

The work done is the integral of P dV. Since PV is constant in an isothermal process, P = P1 V1/V

Work = Integral P1*V1 dV/V

V1=1 to V2=5

= 2.77*10^7 N/m^2 * 10^-3 m^3 * ln 5

= 4.46*10^4 Joules

Check my work. The method is more likely to be correct than my calculations

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