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Work done isothermal, adiabatic ideal gas

  1. Feb 20, 2015 #1
    Problem statement, work done, and relevant equations:

    One mole of ideal gas is initially at 1 atm and has a volume of 5L.

    a) Calculate the work done on the gas during an isothermal, reversible compression to a volume of 2L.

    ##W_isothermal = - \int_{v_i}^{v_f} p dv = - \int_{v_I}^{v_f} \frac{nRT}{V} dV = -nRT \int_{v_I}^{v_f} \frac{1}{V} dV = -nRT ln \frac{V_I}{V_f} = -(1 mol)(8.314 \frac{J}{molK})T ln \frac{.005 m^3}{.002 m^3} = 7.618T##

    b) Calculate the work done during a reversible adiabatic compression to 2L.

    ##P_i = \frac{nRT_I}{V} → T_i = \frac{V_I P_i}{nR} = \frac{(.005 m^3)(101300 Pa)}{(1 mol)(8.314 \frac{J}{molK})} = 60.92 K##

    I think I'm a little stuck here because the problem statement does not give me whether its a monatomic or diatomic gas in order to get the ##\frac{C_p}{C_v}## value. Is there something I'm not seeing?

    c) What are the final pressures for (a) and (b)?
     
  2. jcsd
  3. Feb 21, 2015 #2

    Andrew Mason

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    The first part is correct and your determination of ##T_i## is correct, assuming you have done the arithmetic properly. You do need to know the value for ##\gamma = \frac{C_p}{C_v}## in order to calculate the adiabatic work and the final pressure and temperature (applying the adiabatic condition). I would suggest that you assume a true ideal gas (monatomic).

    AM
     
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