The figure below shows how 3 moles of a monatomic ideal gas follows the threepa
Physics

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For this problem, we are talking about the changes in phases of an ideal gas. We want to find Q, W, and ΔU. When dealing with ideal gases, the ideal gas law applies:
PV = nRT (n is # of moles, R is universal gas constant) Eq. 1
R = 8.314 m^{3}PaK^{1}mol^{1}
Q represents the heat added or expelled from the gas. When Q is positive, heat was added to the gas. When Q is negative, heat was expelled by the gas.
W represents the work done to compress or expand the gas. If a gas doesn’t expand or compress, then Work is zero. When a gas expands or compresses at constant Pressure, the equation for work done is:
W = P ΔV Eq. 2
If W is positive, then the Volume increased so ΔV (change in volume) of the gas is positive (gas expanded). If W is negative, the Volume decreased and ΔV is negative (gas compressed).
If the Pressure is not constant, then we can’t use the equation for Work above. However, if the Temperature is constant, then we can use the following:
W = nRT ln{V_{f}  V_{i}} Eq. 3
ΔU represents the change in U, where U is the internal energy of the gas.
The relationship between these variables is described by this equation:
ΔU = Q – W Eq. 4
The internal energy, U, is also directly related to temperature in this way:
U = nRT Eq. 5
Thus if there’s a change in temperature, then
ΔU = nR ΔT Eq. 6
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So, now that we have all the equations we need in front of us, let’s look at the problem:
n = 3 (3 moles, which doesn’t change throughout the process)
At point A, we are given:
V = 0.0015 m^{3} (volume)
P = 600 kPa = 600,000 Pa (pressure by reading the yaxis of the graph at point A)
We have pressure, volume, and # of moles. So we can use Eq. 1 to find T.
T = PV/(nR) = (600,000)(0.0015)/[(3)(8.314)] = 36.1 K
At point B, we are given:
V = 0.009 m^{3}
P = 100 kPa = 100,000 Pa (read off the yaxis at point B)
T = PV/(nR) = 36.1 K (using Eq. 1)
At point C, we are given:
V = .0015 m^{3} (same xaxis position as point A, so they must have same volume)
P = 100,000 Pa (read off the yaxis)
T = PV/(nR) = 6.0 K (using Eq. 1)
Now let’s look at each process. First, A to B:
It’s given that its isothermal expansion. Isothermal means Temperature is constant, or ΔT = 0. So,
ΔU = 0. This is true for all isothermal processes. (justified by Eq. 6)
To find the Work, we can use Eq. 3, since the Temperature is constant.
W = nRT ln{V_{f}  V_{i}} = (3)(8.314)(36.1)ln{0.009/0.0015} = 1613 Joules
Now using Eq.4, we can find Q,
ΔU = Q – W,
0 = Q – 1613J,
Q = 1613 J
So we are done with A to B. We found ΔU, W, and Q.
Next, B to C:
We can use Eq.6 to find ΔU,
ΔU = nRΔT = nR(T_{C} – T_{B}) = (3)(8.314)(6.0 – 36.1) = 1125 J
Since pressure is constant from B to C, we can use Eq.2 to find W,
W = PΔV = P(V_{C} – V_{B}) = (100,000)(0.0015 – 0.009) = 750 J
Now use Eq.4 to find Q,
Q = ΔU + W = 1125 + (750) = 1875 J
So we are done with B to C.
Finally, C to A:
The volume isn’t changing, therefore
W = 0
Use Eq.6 to find ΔV,
ΔU = nRΔT = nR(T_{A} – T_{c}) = (3)(8.314)(36.1 – 6.0) = 1125 J
Now use Eq.4 to find Q,
Q = ΔU + W = 1125 J + 0 = 1125 J
So we are done with C to A.
That is the full cycle.
The efficiency tells you how much of the heat added was transformed into work.
So, e = W_{out} / Q_{in} is the equation for thermal efficiency.
Q_{in} means all the heat that was put into the gas, or added to it. These are the positive Q’s in the entire cycle added together.
What parts of the cycle had Q positive? From A to B, Q = +1613 J. And from C to A, Q = +1125 J. So,
Q_{in} = 1613 + 1125 = 2738 J
W_{out} means the total work done in the cycle. So add the work done in each part of the cycle (even the negative works, if there are any). So,
W_{out} = 1613 + (750) = 863 J
So, the efficiency is e = 863 / 2738 = 0.315
or 31.5%.
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