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 = p_B

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

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 = nRT_A ∙ln(V_A/V_C)

W = 8(1.013 × 10⁵) ∙ 8.314 ∙ 275K ∙ ln (V_A/V_C)

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 V_A 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 V_A 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

does this answer look alright?

 

[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 T_f my answer is wrong. Am I using the wrong equation?

 

[espen180]

For adiabatic processes, you need the adiabat equation

$$T_1V_1^{\alpha}=T_2V_2^{\alpha}$$ 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]

For adiabatic processes, you need the adiabat equation

$$T_1V_1^{\alpha}=T_2V_2^{\alpha}$$ 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?

Thank you very much, that was very helpful. I really appreciate your help.

 

[espen180]

No problem. :-)