Heat Engines, Heat Capacity of Gas at Constant Pressure

Analyzing the Cycle

A graph of pressure (P) versus volume (V) was sketched for a heat engine. The heat engine has a gas that is heated. The gas expands and does work on the surroundings. Then it is then cooled at constant volume, using a stopper so the piston cannot move, reducing the pressure. The stopper is then removed and the volume decreases to its original position at constant pressure. The gas is heated at constant volume and the pressure rises to the original point and the process is repeated. The first three points were given and we found the rest of the graph information according to the other information given.

From point 1 to 2 the gas expands and volume increases, doing positive work (W+) on the surroundings. The pressure remains constant in this step. There is no heat in this process.

From 2 to 3 the volume is maintained constant and the gas is cooled. Since the volume remains constant, there is no work done. however there is heat loss to the surroundings in this process (Q-).

From 3 to 4 the volume decreases while the pressure remains constant. There is work done on the system (W-) by the surroundings but there is no heat entering or leaving the system.

From 4 to 1 the volume is kept constant while the pressure increases. There is no work done but positive heat (Q+) enters the system.

The internal energy was calculated at each of the four points using the formula E_int = (3/2)PV. We included these energies (in Joules) in our sketch next to each point in red.

The Carnot Engine Cycle

Here we see a repeating process that includes isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic expansion.

We calculate the internal energy at each point by using the equation E_int = (3/2)PV. We are to calculate the heat and work for each of the parts of the carnot cycle (pending).

ActivPhysics Simulations

In this first simulation (problem 6) we are calculating the heat capacity of a gas at constant pressure. We assume 1 mole of gas is in the cylinder.

 The temperature in this simulation starts at 200K with a constant pressure of 166 kPa and volume of 10 dm^3.

We let the simulation run until the temperature reaches 401K. At this point we can see the changes in volume, temperature, heat, work, and internal energy. We divide the change in heat (Q_f - Q_i) by the change in temperature (T_f - T_i) and by the 1 mole (n) of gas in the cylinder: (Q_f - Q_i)/((T_f - T_i)(n)). This gives us a 20.9 J/(mol K).

We ran the simulation from the previous point and stopped it immediately. This gave us a heat change of 354 J and a temperature change of 18 K. Using the same formula in the previous calculation, we calculated the specific heat of the gas at constant pressure and found it to be 19.7 J/(mol K).

Again we ran the simulation and stopped it. We calculated for heat capacity again and got 20.9 J/(mol K).

We repeated the same steps one last time and got the heat capacity of the gas at constant pressure to be 20.8 J/(mol K).

Averaging the values for the four trials gives us 20.6 J/(mol K). This comes very close to the actual value 20.4 J/(mol K).

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In this simulation (problem 7), we started the simulation at an initial temperature of 200K and stopped it at 797K.

We did the same as the in the previous calculations for heat capacity of a gas at constant temperature. This time we got 20.8 J/(mol K). This matches the 20.8 J/(mol K) we were supposed to get.

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On this whiteboard, we derive the specific heat of an ideal gas at constant pressure (problem 8). We see that we get the same 20.8 J/(mol K) that we obtained from the simulations in the previous problem.