GSO001 Problem 14

The internal energy $U$U of a monatomic ideal gas is:

$$U = \frac{3}{2}nRT = \frac{3}{2}PV$$U=23​nRT=23​PV

1. State Variables at Each State

• State A: Pressure $P_0$P0​, volume $V_0$V0​, internal energy $U_A = \frac{3}{2}P_0V_0$UA​=23​P0​V0​
• State C (isochoric A $\to$→ C): Pressure $2P_0$2P0​, volume $V_0$V0​, internal energy $U_C = \frac{3}{2}(2P_0)(V_0) = 3P_0V_0$UC​=23​(2P0​)(V0​)=3P0​V0​
• State B (isobaric C $\to$→ B): Pressure $2P_0$2P0​, volume $2V_0$2V0​, internal energy $U_B = \frac{3}{2}(2P_0)(2V_0) = 6P_0V_0$UB​=23​(2P0​)(2V0​)=6P0​V0​

2. Total Heat Absorbed $Q_{in}$Qin​

Using the first law $Q = \Delta U + W$Q=ΔU+W:

• Process A $\to$→ C (isochoric heating): $W_{AC} = 0$WAC​=0

$$Q_{AC} = U_C - U_A = 3P_0V_0 - \frac{3}{2}P_0V_0 = \frac{3}{2}P_0V_0$$QAC​=UC​−UA​=3P0​V0​−23​P0​V0​=23​P0​V0​

This is positive, so heat is absorbed.

• Process C $\to$→ B (isobaric heating): $W_{CB} = 2P_0 \times (2V_0 - V_0) = 2P_0V_0$WCB​=2P0​×(2V0​−V0​)=2P0​V0​

$$Q_{CB} = (U_B - U_C) + W_{CB} = (6P_0V_0 - 3P_0V_0) + 2P_0V_0 = 5P_0V_0$$QCB​=(UB​−UC​)+WCB​=(6P0​V0​−3P0​V0​)+2P0​V0​=5P0​V0​

This is also positive, so heat is absorbed.

Note: Process B $\to$→ A releases heat ($Q_{BA} = -6P_0V_0$QBA​=−6P0​V0​) as stated in the problem. This is not included in the denominator of thermal efficiency.

Total heat absorbed:

$$Q_{in} = Q_{AC} + Q_{CB} = 1.5P_0V_0 + 5P_0V_0 = 6.5P_0V_0 = \frac{13}{2}P_0V_0$$Qin​=QAC​+QCB​=1.5P0​V0​+5P0​V0​=6.5P0​V0​=213​P0​V0​

3. Net Work $W_{net}$Wnet​

The net work per cycle equals the area of the closed loop on the $P$P-$V$V diagram. The vertices are $(V_0, P_0)$(V0​,P0​), $(V_0, 2P_0)$(V0​,2P0​), $(2V_0, 2P_0)$(2V0​,2P0​), forming a right triangle with base $V_0$V0​ and height $P_0$P0​:

$$W_{net} = \frac{1}{2} \times V_0 \times P_0 = \frac{1}{2}P_0V_0$$Wnet​=21​×V0​×P0​=21​P0​V0​

4. Thermal Efficiency

$$e = \frac{W_{net}}{Q_{in}} = \frac{\frac{1}{2}P_0V_0}{\frac{13}{2}P_0V_0} = \frac{1}{13}$$e=Qin​Wnet​​=213​P0​V0​21​P0​V0​​=131​

Note: $P_0V_0$P0​V0​ cancels in the calculation, so the specific numerical values given ($V_0 = 1.0 \times 10^{-3}\ \mathrm{m^3}$V0​=1.0×10−3 m3, $P_0 = 1.0 \times 10^5\ \mathrm{Pa}$P0​=1.0×105 Pa) are not needed.

With $e = \frac{1}{13}$e=131​: $X = 1$X=1, $Y = 13$Y=13.

The required value is $X + Y = 1 + 13 = 14$X+Y=1+13=14.

Answer: 14