Thermodynamic Cycle 2: Understanding PV^y, Cp/Cv and PV=nRT

AI Thread Summary
The discussion focuses on applying thermodynamic principles to solve problems related to the PV^y relationship, specifically in the context of a thermodynamic cycle. Participants are trying to relate variables such as Vm and Vo using the equation PV^y = constant, and confirming calculations for pressure ratios. They also discuss determining work done, heat transfer, and changes in internal energy for an isobaric process, utilizing the first law of thermodynamics. The conversation highlights the challenge of finding temperature changes and the need to express variables per mole for clarity. The molar heat capacity for a monoatomic gas is confirmed as Cv = 3/2 R, aiding in the calculations.
subzero0137
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2. PV^y = const, y=Cp/Cv, PV=nRT, 3.
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I've drawn the cycle for part a) of the question, but I'm not sure how to do part b). I know I need to relate Vm and 4Vo using the PV^y = constant relation but I'm stuck as to how to do it.
 
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subzero0137 said:
I know I need to relate Vm and 4Vo using the PV^y = constant relation
Using that, write an equation relating P0, V0, Pm and Vm.
 
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haruspex said:
Using that, write an equation relating P0, V0, Pm and Vm.

Thanks, I think I got it now. Just to confirm, would P_m = P_0/32?
 
subzero0137 said:
Thanks, I think I got it now. Just to confirm, would P_m = P_0/32?
Yes.
 
haruspex said:
Yes.

Sorry to bother you again but the next part of the questions asks to determine the work done, heat transfer and change in internal energy in each process. So if we consider the first, isobaric process then the work done on the gas is simply Won = -P0 * (4V0 - V0) = -P0 * 3V0. To calculate the heat transferred (Q) and change in internal energy (ΔU) I can use the first law of thermodynamics, but the equations for Q and ΔU involve variables that I don't know. ΔU=nCVΔT and Q=nCPΔT. I don't know how to get either Q or ΔU if I don't know n (moles) and change in temperature.
 
subzero0137 said:
Sorry to bother you again but the next part of the questions asks to determine the work done, heat transfer and change in internal energy in each process. So if we consider the first, isobaric process then the work done on the gas is simply Won = -P0 * (4V0 - V0) = -P0 * 3V0. To calculate the heat transferred (Q) and change in internal energy (ΔU) I can use the first law of thermodynamics, but the equations for Q and ΔU involve variables that I don't know. ΔU=nCVΔT and Q=nCPΔT. I don't know how to get either Q or ΔU if I don't know n (moles) and change in temperature.
Try it algebraically, also making use of the ideal gas law, and see how it plays out. I think you will be pleasantly surprised.
 
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Chestermiller said:
Do it "per mole."

ΔU=CVΔT and Q=CPΔT? But the temperature differences are still missing. I tried to do P0V0/T1 = P04V0/T2 ⇒ T2=4T1 but that doesn't give a change in temperature
 
$$nT_0=\frac{P_0V_0}{R}$$
$$nT_1=\frac{P_0(4V_0)}{R}$$
$$nC_v(T_1-T_0)=C_v\frac{3P_0V_0}{R}$$
For a monoatomic gas, what is the molar Cv in terms of R?
 
Chestermiller said:
$$nT_0=\frac{P_0V_0}{R}$$
$$nT_1=\frac{P_0(4V_0)}{R}$$
$$nC_v(T_1-T_0)=C_v\frac{3P_0V_0}{R}$$
For a monoatomic gas, what is the molar Cv in terms of R?

Molar Cv = 3/2 R? Sorry I'm still confused
 
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subzero0137 said:
Molar Cv = 3/2 R? Sorry I'm still confused
$$\Delta U=nC_v\Delta T=\left(\frac{3R}{2}\right)\frac{3P_0V_0}{R}=?$$
 
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