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An ideal gas initially at pressure P0, volume V0, and temperature T0 is taken through the cycle described in the figure below. (Assume n = 4 and m = 2.)A rectangular path is plotted on a PV diagram that has a horizontal axis labeled V, and a vertical axis labeled P. The area inside the rectangle is shaded. The path is clockwise and runs through corner points A–D in the following order:A (V0, P0), B (V0, nP0), C (mV0, nP0), D (mV0, P0).(a) Find the net work done by the gas per cycle in terms of P0 and V0. (Substitute numeric values for n and m, do not use the variables n and m.)Wenv = 3 (b) What is the net energy Q added to the system per cycle? (Use the following as necessary: P0 and V0.)Q = 3 (c) Obtain a numerical value for the net work done per cycle for 1.00 mol of gas initially at 0°C. Hint: Recall that the work done by the system equals the area under a PV curve. kJ

Question

An ideal gas initially at pressure P0, volume V0, and temperature T0 is taken through the cycle described in the figure below. (Assume n = 4 and m = 2.)A rectangular path is plotted on a PV diagram that has a horizontal axis labeled V, and a vertical axis labeled P. The area inside the rectangle is shaded. The path is clockwise and runs through corner points A–D in the following order:A (V0, P0), B (V0, nP0), C (mV0, nP0), D (mV0, P0).(a) Find the net work done by the gas per cycle in terms of P0 and V0. (Substitute numeric values for n and m, do not use the variables n and m.)Wenv = 3 (b) What is the net energy Q added to the system per cycle? (Use the following as necessary: P0 and V0.)Q = 3 (c) Obtain a numerical value for the net work done per cycle for 1.00 mol of gas initially at 0°C. Hint: Recall that the work done by the system equals the area under a PV curve. kJ

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Solution

(a) The work done by the gas in a cycle is given by the area enclosed by the cycle in the PV diagram. In this case, the cycle is a rectangle with sides of length (nP0 - P0) and (mV0 - V0). Therefore, the work done is:

W = (nP0 - P0) * (mV0 - V0)

Substituting the given values of n = 4 and m = 2, we get:

W = (4P0 - P0) * (2V0 - V0) = 3P0V0

(b) For an ideal gas in a cyclic process, the net heat added to the system is equal to the net work done by the gas. Therefore, Q = W = 3P0V0.

(c) To find the numerical value of the work done, we need to express P0 and V0 in terms of the given initial conditions. For 1.00 mol of gas at 0°C, we can use the ideal gas law P0V0 = nRT0, where R is the gas constant (8.314 J/(molK)) and T0 is the initial temperature in Kelvin (273.15 K). Therefore, P0V0 = (1.00 mol)(8.314 J/(molK))(273.15 K) = 2271.3 J.

Substituting this into the expression for the work done, we get:

W = 3P0V0 = 3 * 2271.3 J = 6813.9 J = 6.8139 kJ.

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