# Work in thermodynamic processes

roam

## Homework Statement

Two moles of an ideal monoatomic gas trebles its initial volume in an isobaric expansion from state A to state B. The gas is then cooled isochorically to state C and finally compressed isothermally until it returns to state A. The molar gas constant is R = 8.314 J mol–1 K–1 and Boltzmann's constant is 1.38 × 10–23 J K–1.

State B corresponds to a pressure P = 8 atm (1 atm = 1.013 × 105 Pa) and temperature T = 552°C. (I also know that temperature of the gas in state A is 275 K)

What is the work done on the surroundings in each of the following processes:

(i) A→B

(ii) C→A

## The Attempt at a Solution

(i) This process is isobaric
p = constant = pB

$$W=-P(V_B-V_A)$$

I could calculate the volumes from the given information. But it easier to substitute pV = nRT for each state and get:
$$W = nR(T_B - T_A)$$
W = 8(1.013 × 105) ∙ (8.314).(825.15 K - 275.05 K)
W= 3706389847 J

Why is the value for work so large? The correct answer to this part must be 9150.

(ii)
Since we have an isothermal process:
$$pV = nRT = constant = nRT_C$$

W = nRTA ∙ln(VA/VC)

W = 8(1.013 × 105) ∙ 8.314 ∙ 275K ∙ ln (VA/VC)

I'm stuck. How do I get the volumes for states A and C?

(P.S. I also know that the work done in the processe B→C is 0, because it is an isochoric (i.e. isovolumetric) process).

espen180
You can find VA directly from the given information, as you yourself stated in (i).

Next is state C. The gas is cooled isochorically from B to C. What does that tell you?

roam
You can find VA directly from the given information, as you yourself stated in (i).

Next is state C. The gas is cooled isochorically from B to C. What does that tell you?

It tells me that the volume of state C is equal to the volume of state B. So I just need to find the volume for A and B(=C).

So for the volume of C:

$$\frac{PV}{T}=nR$$

$$\frac{8(1.013 \times 10^5)V}{825 K}=n(8.314)$$

What value do I need to use for the number of moles?

espen180
Look through the problem statement.

roam
Look through the problem statement.

I already have. If I use 8 (1.013 × 105) for the pressure, I can't use it again for the number of moles!

espen180
What do the two first words say?

roam
What do the two first words say?

Wow, that was dumb of me. I'm trying to do a lot of things at the same time :shy:
I managed to get part (i) right using the right value for n.

But for part (ii):

8(1.013 × 105)V=2(8.314)(825 K)

V=0.016

Volume of C is 0.016. And the problem says "gas trebles its initial volume in an isobaric expansion from state A to state B", so that means the volume at C is 3 times larger than that of A:

$$V_A=\frac{V_C}{3} = \frac{0.016}{3}=5.3 \times 10^{-3}$$

Are these values correct?

W = 2 ∙ 8.314 ∙ 275K ∙ ln (5.3 \times 10^{-3}/0.016)

W=-5052.29

espen180
I haven't done the numeric calculation myself, but your method checks out.

Efficiency tip: Since you know the gas triples in volume, you don't really need to calculate the volume of before/after for the isothermal compression. You know that

$$w=nRT\ln\left(\frac{V_A}{V_C}\right)=nRT\ln\left(\frac{V_A}{3V_A}\right)=-nRT\ln 3$$

roam
Thanks a lot for the tips!

This is my last question: The gas in state A now expands adiabatically to 7 times its original volume (State D). What is the final temperature of the gas?

I tried to use the equation of state for an ideal gas PV=nRT to find the temprature:

$$\frac{PV}{T}=nR$$

$$\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$$

I know that in an adiabatic process Q=0 meaning that no energy enters or leaves the system by heat. We don't know whether or not the pressure is constant, but since I don't know the final pressure I assume it is constant

$$\frac{1}{T_i}=\frac{7}{T_f}$$

When I substitute the values and solve for the Tf my answer is wrong. Am I using the wrong equation?

espen180

$$T_1V_1^{\alpha}=T_2V_2^{\alpha}$$ where $$\alpha$$ is the number of degrees of freedom divided by 2, which is $$\alpha=\frac{3}{2}$$ for a monoatomic ideal gas, or $$\alpha=\frac{5}{2}$$ for a diatomic ideal gas.

Can you take it from there?

roam
$$T_1V_1^{\alpha}=T_2V_2^{\alpha}$$ where $$\alpha$$ is the number of degrees of freedom divided by 2, which is $$\alpha=\frac{3}{2}$$ for a monoatomic ideal gas, or $$\alpha=\frac{5}{2}$$ for a diatomic ideal gas.