One mole of an ideal gas at a pressure of 1 atm and temperature of 273 K is expanded to twice its volume at constant pressure. a. In terms of P and V 1 and V 2 how much work was performed? Was the work performed on the gas or by the gas? Explain. b. If two moles of an ideal gas at the same initial

Question

One mole of an ideal gas at a pressure of 1 atm and temperature of 273 K is expanded to twice its volume at constant pressure.
a. In terms of P and V_1 and V_2 how much work was performed? Was the work performed on the gas or by the gas? Explain.
b. If two moles of an ideal gas at the same initial pressure and temperature were to double its volume under constant pressure, how much work would be performed on or by the gas? Compare the value with that in part (a).
c. In either case (a) or (b), does the temperature increase or decrease? Explain.

Accepted Answer

a. The work is performed by the gas and equals [latex] P Δ V = P (V_2 – V_1 ) = P V_1 [/latex] .
b. The starting volume for this part of the problem is [latex] 2V_1 = V_2 [/latex]. This doubles to [latex]4V_1 [/latex] .
Thus, the work performed by 2 moles of the gas is [latex] 2PV_1 [/latex] , twice that of part (a) of the problem.
c. The temperature increases in both cases. Take case (a):
[latex]P_1 V_1 = RT_1 and P_2 V_2 = RT_2 [/latex] , but [latex] P_2 = P_1 and V_2 = 2V_1 [/latex].
Thus, [latex] T_2 = 2T_1 [/latex]. The temperature increases.
Why does the temperature increase? In order to expand at constant pressure, more energy must be given to the gas to keep the pressure from decreasing.

Question 2.12.1: One mole of an ideal gas at a pressure of 1 atm and temperat...

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