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.
C to A is a compression, so negative work is done by the gas.
The relationship between P and V during the isothermal compression from C to A is
P (atm) = 2 (liter*atm)/V (liter)
The work done ON the gas in compressing from C to A is
Integral of P dV = Intergsl of 2 L*atm dV/V
V 5 to 1
= 2 L*atm*ln 5 = 3.219 L*atm
3.219*10^-3 m^3*1.013*10^5 N/m^2 = 326 J
The negative of that value is the work done by the has.
You do not need to use the 480 complete cycle work number they provided. That is calculable directly from a P-V plot. I got 488 J
To find the work done by the gas during process C to A, we need to use the ideal gas law and the given information about the temperature and volume.
First, let's find the initial pressure at point C 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. We are given V = 5 L and T = 300 K. Since the process C to A is isothermal, the temperature remains constant at 300 K.
Let's assume one mole of gas for simplicity. Rearranging the equation, we get:
P = (nRT) / V
P = (1 mol * 0.0821 L atm/mol K * 300 K) / 5 L
P = 4.926 atm
Now we can find the work done during process C to A using the formula:
W = P∆V
Since the process is compression, ∆V = Vfinal - Vinitial. Rearranging the equation, we get:
W = P(Vinitial - Vfinal)
W = 4.926 atm * (1 L - 5 L)
W = -19.704 atm L
Converting atm L to joules using the conversion factor 1 atm L = 101.325 J, we get:
W = -19.704 atm L * 101.325 J / 1 atm L
W = -1992 J
Therefore, the work done by the gas during process C to A is -1992 J.