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
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 ,
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