1) from an initial state of the gas described by T = 14.5degrees C and P = 2.00

Question

1) from an initial state of the gas described by T = 14.5degrees C and P = 2.00 x 104 Pa, the gas undergoes anisothermal expansion against a constant external pressure of 1.00 x104 Pa until the volume has doubled.
2) subsequently the gas is cooled at constant volume. Thetemperature falls to -35.6 degrees C.

Explanation / Answer

1) isothermal => no temperature change => no change in U and Hb/c they are state functions of T V_initial = nRT/P = (1.65 mol *8.314 J/mol/K *287.5 K)/(2.00e4 Pa)= 0.197197688 m3 V_final = 2*V_initial W = -Pext*V = -Pext*(V_final - V_initial) =-Pext*(2V_initial- V_initial) = -Pext*V_init W = -1.00e4 Pa*(0.187187688 m3) = -1.97 e3 J = -1.97 kJ  (system does 1.97 kJ on environment) U = q + W = 0   (recall isothermal, so nochange in energy) q = -W = 1.97 kJ 2) U = n*Cv*T = 1.65 mol *[(3*8.314 J/mol/K)/2]*(-35.6 -14.5) = -1030.91521 J = -1.03 kJ W = 0 b/c V is constant U = q + W = q = -1.03 kJ For an ideal gas, Cp = Cv + R = 5R/2 H = n*Cp*T = 1.65 mol *[(5*8.314 J/mol/K)/2]*(-35.6 -14.5) = -1718.19203 J = -1.72 kJ

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