How Do You Calculate State Changes in a Monatomic Gas Thermodynamic Cycle?

[rchenkl]

Homework Statement

Five moles of an ideal monotomic gas initially occupies a volume of 100 x 10^-3m³ at a temperature of 280 K. (state 1)
The gas is then subject to the following processes in sequence:
* heated at constant volume to a temperature of 600K (state 2)
* allowed to expand isothermally to its initial pressure (state 3)
* compressed isobarically to its original volume (state 1)

(a) Find the pressure and volume of the gas in state 3
(b) Calculate the work done on the gas in going from state2 to state 3.
(c) Calculate the heat exchanged between the gas and its environment during each of the 3 processes of the cycle 1->2, 2->3, 3->1.
In each case, indicate whether the heat enters or leaves the gas.
(d) Calculate the net work done on the gas in one cycle

Homework Equations

PV=nRT=Nk_BT
pV^r=constant, r=C_p/C_v
dW=-PdV
W=nRT*ln(V_i/V_f)
E_int=3/2Nk_bT
C_v=5/2R

The Attempt at a Solution

I'm stuck on question (a).
I calculated the pressure using PV=nRT where the initial volume, temperature and number of moles of gas is provided. P=(5*8.314*280)/(100x10^-3)
Now, I don't know how to calculate the volume in state 3. Do I use T=600K and P=the above I calculated, to obtain the volume?

 

[Mapes]

You got it.

 

[nlightner]

As a scientist, it is important to first make sure that all units are consistent and in the correct SI units. In this problem, the volume is given in m3 and the temperature in Kelvin, but the pressure is given in units of 10-3m3. This should be converted to Pa (Pascal) by multiplying by 1000.

For part (a), the ideal gas law can be used to solve for the volume in state 3. Since the number of moles and temperature remain constant, the equation becomes P1V1 = P2V2. Using the initial values for state 1 and the given pressure for state 3, the volume can be calculated as follows:

P1V1 = P2V2
(5*8.314*280)/(100x10^-3) = (P2)(V2)
V2 = (5*8.314*280)/(100x10^-3*P2)

For part (b), the work done on the gas can be calculated using the equation W = nRT*ln(Vi/Vf). In this case, n and T remain constant, so the work done is directly proportional to the natural log of the initial and final volumes.

For part (c), the heat exchanged between the gas and its environment can be calculated using 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. In this problem, the internal energy remains constant since the temperature remains constant, so the heat exchanged can be calculated as follows:

Q = W + ΔEint
Q = W + 0
Q = W

Since the work done was calculated in part (b), the heat exchanged during each process can be determined by looking at the sign of the work done. If the work done is positive, then heat is entering the system, and if the work done is negative, then heat is leaving the system.

For part (d), the net work done on the gas in one cycle can be calculated by adding the work done in each process. Since the gas returns to its initial state, the net work done should be equal to zero. If it is not, then there may be an error in the calculations.