A heat engine using a monatomic gas follows the cycle shown in the pV diagram to the right. The gas starts out at point 1 with a volume of 233 cm3, a pressure of 235 kPa, and a temperature of 317 K. The gas is held at a constant volume while it is heated until its temperature reaches 395 K (point 2). The gas is then allowed to expand adiabatically until its pressure is again 235 kPa (point 3). The gas is maintained at this pressure while it is cooled back to its original temperature of 317 K (point 1 again).
For each stage of this process, calculate in joules the heat Q(in) transferred to the gas, and the work W(out) done by the gas.
(Stages)
Q(in) W(out)
1 > 2
2 > 3
3 > 1
Q(in) = 5,845 J
W(out) = -5,845 J
To calculate the heat transferred to the gas (Qin) and the work done by the gas (Wout) for each stage of the process, we can use the equations for heat and work in a thermodynamic process.
1 > 2:
This stage involves heating the gas at a constant volume. Since the volume is constant, the work done is zero (Wout = 0). The heat transferred is given by the equation: Q = nCv(T2 - T1), where n is the number of moles of gas and Cv is the molar specific heat capacity at constant volume.
To calculate the value of Qin, we need to know the molar specific heat capacity at constant volume for a monatomic gas. The molar specific heat capacity at constant volume for a monatomic gas is Cv = (3/2)R, where R is the ideal gas constant.
2 > 3:
This stage involves an adiabatic expansion where no heat is transferred to the gas (Qin = 0). The work done can be calculated using the equation: W = nCp(T3 - T2), where Cp is the molar specific heat capacity at constant pressure. For a monatomic gas, the molar specific heat capacity at constant pressure is Cp = (5/2)R.
3 > 1:
This stage involves cooling the gas at a constant pressure. Again, the work done is zero (Wout = 0). The heat transferred can be calculated using the equation: Q = nCv(T1 - T3).
To calculate the specific values of Q(in) and W(out), we need to know the number of moles of gas (n) and the values of the temperatures (T1, T2, T3).
To calculate the heat (Qin) transferred to the gas and the work (Wout) done by the gas for each stage of the process, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
1) From point 1 to point 2:
Since the gas is held at constant volume, the work done by the gas is zero (Wout = 0). Therefore, we only need to calculate the heat transferred to the gas (Qin).
The change in internal energy (ΔU) can be calculated using the equation ΔU = Q - W, where ΔU = 3/2 nRΔT (for monatomic gas) and nR = Cv (molar specific heat at constant volume).
So, ΔU = Cv ΔT = Cv (395 K - 317 K).
Now, assuming the process is reversible, we know that Qin = ΔU + Wout (since Wout = 0), so Qin = Cv (395 K - 317 K).
2) From point 2 to point 3:
In an adiabatic process, there is no heat exchange with the surroundings (Qin = 0). Therefore, we only need to calculate the work done by the gas (Wout).
The work done in an adiabatic process can be calculated using the equation Wout = ΔU = CvdT, where CvdT = (3/2 - R) nΔT.
Therefore, Wout = (3/2 - R) n (395 K - 317 K).
3) From point 3 to point 1:
Again, the gas is held at constant volume, so the work done by the gas is zero (Wout = 0). Therefore, we only need to calculate the heat transferred to the gas (Qin).
Using the same reasoning as in point 1, Qin = Cv (317 K - 395 K).
Now, to calculate the values of Qin and Wout, we need to know the specific heat capacity at constant volume (Cv) and the gas constant (R) for the given monatomic gas. These values can be looked up in a thermodynamics reference table or provided in the question.
Once the specific values of Cv and R are known, substitute them into the equations to obtain the values of Qin and Wout for each stage of the process.