One mole of an ideal monatomic gas is taken through the cycle shown in the figur

Question

One mole of an ideal monatomic gas is taken through the cycle shown in the figure. Assume that p = 4 p0, V = 4 V0, p0= 1.01

Explanation / Answer

1. Use ideal gas law p·V = n·R·T p·V/T = n·R = constant , because n is constant Therefore pa·Va/Ta = pb·Vb/Tb => Tb = Ta · (pb/pa) · (Vb/Va) from the graph you can see that Vb=Vc and pb=pa. Hence: Tb = Ta · (Vc/Va) = 259K · (0.0525m³ / 0.0201) = 676K 2. The change of enthalpy is equal to the heat transferred fro a constant pressure process like a?b. ?Hab = Qab For an ideal gas enthalpy is a function of temperature alone: H = n·Cp·T (Cp is the greater of the two given heat capacities. This is because Cp - Cv = R for ideal gases) Hence: Qab = ?Hab = n·Cp·(Tb - Ta) = 1mol · 20.8J/mol · (676K - 259) = 8673.6J 3. Like Enthalpy internal energy is a function of temperature , given by: U = n·Cv·T => ?Uab = n·Cv·(Tb - Ta) = 1mol · 12.5J/mol · (676K - 259) = 5212.5J 4. You can find the work done on the gas from the integral W = - ? p dV from initial to final state. For process like b?c, in which the volume does not change, no work is done. Therefore: Wbc = 0 5. Internal energy is state function, Returning to state a at the end of cycle, means returning to initial Internal energy. So the net change of internal energy in the whole cycle is zero. ?U_net = 0 On the other hAnd the net work done on the gas and the heat transferred to the gas sum up the change of Internal energy ?U_net = Q_net + W_net Hence: W_net = -Q_net = -1400J Negative sign indicates that work is done BY the gas to the surrounding. 6. The efficiency of a carnot engine is given by: ?_carnot = 1 - T_cold/T_hot with T_hot = Tb and T_cold = Ta=Tc ?_carnot = 1 - 259K/676K = 0.617 = 61.7% Actual efficiency of the process would be net work done divided by heat transferred to gas: ? = -W_net / Qab = 1400 / 8673.6 = 16.1%

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