Post a New Question


posted by .

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 -

    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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question