A heat engine has the three step cycle shown above. Starting from point A, 1 L (liter) of ideal gas expands in the isobaric (constant pressure) process A to B at P = 2 atm. B to C is isovolumetric process at V = 5 L, and C to A is an isothermal (constant temperature) compression at 300 K. (Note that the diagram is illustrating V vs P. This should make no difference in your calculations. )
i found the pressure at point C to be .4 atm and the heat exhausted from B to C to be 6000J
also on the picture it shows from C to A T=300[k]
If the net work done by the engine per cycle is 480 J, find the work done by the gas during process C to A??.
i got -5681 but that isn't right
* Physics - Chris, Friday, May 7, 2010 at 8:29pm
pA = pB = 2 atm = 2 * 101325 Pa = 202650 Pa
vA = 1 L = 1 * 10^-3 m^3
vB = 5 L = 5 * 10^-3 m^3
W(A to B) = 202650 * (5 * 10^-3 - 1 * 10^-3) = 202650 * 4 * 10^-3 = 810.6 J
W(B to C) = 0 (because volume is constant)
W(in complete cycle) = 480 J
W(in complete cycle) = W(A to B) + W(B to C) + W(C to A)
480 = 810.6 + 0 + W(C to A)
W(C to A) = 480 - 810.6 = -330.6 J
Ans: -330.6 J
That isn't the right answer either.
To find the work done by the gas during process C to A, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added (Q) minus the work done (W) on the system.
ΔU = Q - W
We know that process C to A is an isothermal process, which means the temperature does not change. Therefore, the change in internal energy (ΔU) is zero.
ΔU = 0
Therefore, we can rewrite the equation as:
0 = Q - W
We are given that the net work done by the engine per cycle is 480 J. Since the engine is a heat engine, the net work done is equal to the heat added (Q) minus the work done by the gas (W).
480 = Q - W
We are also given that the heat exhausted from B to C is 6000 J. Since process B to C is isovolumetric (constant volume), no work is done by the gas. Therefore, all the heat exhausted is equal to the heat added.
Q = 6000 J
Substituting the values into the equation, we have:
480 = 6000 - W
Rearranging the equation, we can solve for W:
W = 6000 - 480
W = 5520 J
Therefore, the work done by the gas during process C to A is 5520 J.
To find the work done by the gas during process C to A, we can use the formula:
W(C to A) = -Q(C to A)
Where Q(C to A) is the heat transferred during process C to A.
Since process C to A is an isothermal process, the heat transferred can be calculated using the formula:
Q(C to A) = nRT ln(V_A/V_C)
Where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, ln is the natural logarithm, V_A is the volume at point A, and V_C is the volume at point C.
Given that the volume at point A (V_A) is 1 L = 1 * 10^-3 m^3, and the volume at point C (V_C) is 5 L = 5 * 10^-3 m^3, and the gas is ideal, we can calculate the number of moles using the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
At point A, the pressure (P_A) is 2 atm = 2 * 101325 Pa = 202650 Pa, and the temperature (T) is 300 K.
Plugging in the values, we can solve for n:
202650 * 1 * 10^-3 = n * 8.314 * 300
n = (202650 * 1 * 10^-3) / (8.314 * 300)
n ≈ 0.0819 moles
Now, we can calculate the heat transferred using the formula mentioned earlier:
Q(C to A) = 0.0819 * 8.314 * 300 * ln(1 * 10^-3 / 5 * 10^-3)
Q(C to A) ≈ -248.38 J
Finally, we can find the work done by the gas during process C to A:
W(C to A) = -Q(C to A)
W(C to A) ≈ -(-248.38)
W(C to A) ≈ 248.38 J
So, the work done by the gas during process C to A is approximately 248.38 J.