Question:

An ideal gas is taken through a cycle consisting of four processes, as shown in the following figure:

The gas initially undergoes a constant volume process (Process A-B) where its temperature increases from T1 to T2. Next, the gas expands isothermally (Process B-C) at temperature T2, resulting in a decrease in pressure from P1 to P2. After that, the gas undergoes an adiabatic expansion (Process C-D) where the temperature decreases to T3. Finally, the gas is compressed isothermally (Process D-A) at temperature T3, leading to an increase in pressure from P2 to P1.

1. In which process(es) is work done on the gas? a. A-B b. B-C c. C-D d. D-A e. None of the above

2. In which process(es) is heat supplied to the gas? a. A-B b. B-C c. C-D d. D-A e. None of the above

3. Determine the net work done by the gas during one complete cycle.

Answer:

1. The work done on an ideal gas is given by the equation:

   

   The process where work is done on the gas is when the gas undergoes compression, i.e., when the volume decreases. Looking at the given diagram, we can see that this occurs in Process D-A. Therefore, the answer is d. D-A.

2. The heat supplied to an ideal gas is given by the equation:

   

   Where ΔU represents the change in internal energy, and W represents the work done on the gas.

   Since internal energy only depends on the change in temperature and not on volume or pressure, the only process that involves a change in temperature is Process A-B. Therefore, the heat is supplied to the gas in Process A-B. The answer is a. A-B.

3. The net work done by the gas during one complete cycle is given by the sum of the work done in each individual process.

   Let's calculate the work done in each process:

   • In Process A-B, the volume remains constant, so the work done is zero.

   • In Process B-C, the process is isothermal (constant temperature), and the equation for work done is:

     

     Where n is the number of moles of the gas, R is the ideal gas constant, T is the temperature, V2 is the final volume, and V1 is the initial volume. However, since T2 = T1, the logarithmic term becomes zero, so the work done is also zero.

   • In Process C-D, the process is adiabatic (no heat exchange), and the equation for work done is:

     

     Where γ is the heat capacity ratio. However, since it is not given, we cannot calculate the exact work done in this process.

   • In Process D-A, the process is isothermal (constant temperature), and the equation for work done is:

     

     Where n is the number of moles of the gas, R is the ideal gas constant, T is the temperature, V1 is the initial volume, and V2 is the final volume.

   Since the volumes are reversed in Process D-A, the work done will be the negative of the work done in Process B-C (which is zero). Therefore, the net work done by the gas during one complete cycle is zero.

   The answer is 0 J.