How Is Work and Heat Calculated in a Monoatomic Gas Heat Engine Cycle?

[jincy34]

Homework Statement

A heat engine using a monoatomic gas follows the cycle shown in Figure. Part 2®3 is adiabat, Vi=101 cm3, Vf=668 cm3, Pi=118 kPa, T1=315 K.
Find Q for process 1 to 2.
Find Ws for process 2 to 3.

Homework Equations

PV=nRT
Q=nCv(delta T)

The Attempt at a Solution

I found moles to be 0.0045. But, I don't know where to go from there. I don't know how to find the tempertaure for step 2 and 3.
please help

the picture is added as an attachment.

 

[anathema]

Hello,

Thank you for your post. From the information given, we can use the ideal gas law (PV=nRT) to calculate the temperature for step 2 and 3. We know the initial volume (Vi=101 cm3) and pressure (Pi=118 kPa) for step 2, so we can rearrange the ideal gas law to solve for the temperature (T2) at this step:

T2 = (PiVi)/(nR)

Substituting in the values we have, we get:

T2 = (118 kPa * 101 cm3)/(0.0045 mol * 8.31 J/mol*K)

T2 = 277.6 K

Similarly, for step 3, we know the final volume (Vf=668 cm3) and pressure (Pi=118 kPa), so we can solve for the temperature (T3) at this step:

T3 = (PiVf)/(nR)

Substituting in the values we have, we get:

T3 = (118 kPa * 668 cm3)/(0.0045 mol * 8.31 J/mol*K)

T3 = 1842.2 K

Now, we can use the equation Q=nCv(delta T) to calculate the heat (Q) for process 1 to 2. We know the initial and final temperatures (T1=315 K and T2=277.6 K) and the number of moles (n=0.0045 mol), so we can calculate the heat as follows:

Q = 0.0045 mol * 3/2 * 8.31 J/mol*K * (277.6 K - 315 K)

Q = -0.149 J

Note that the negative sign indicates that heat is being lost from the system in this process.

Finally, to find the work (Ws) for process 2 to 3, we can use the equation Ws = nRTln(Vf/Vi). We know the temperature (T2=277.6 K) and the initial and final volumes (Vi=101 cm3 and Vf=668 cm3), so we can calculate the work as follows:

Ws = 0.0045 mol * 8.31 J/mol*K * ln(668 cm3/101 cm3)

Ws =

 

[ogb p]

I would suggest using the ideal gas law equation PV=nRT to solve for the temperature at steps 2 and 3. From the given information, we know the initial volume (Vi), final volume (Vf), initial pressure (Pi), and initial temperature (T1). We also know the number of moles (n) from the given information. Using these values, we can rearrange the ideal gas law equation to solve for the temperature at each step.

For step 2, we have Vi=101 cm^3, Pi=118 kPa, and n=0.0045 moles. Plugging these values into the ideal gas law equation, we can solve for T2:

T2 = (PiVi)/(nR) = (118 kPa)(101 cm^3)/(0.0045 mol)(8.3145 J/mol*K) = 2,619 K

Similarly, for step 3, we have Vf=668 cm^3, Pi=118 kPa, and n=0.0045 moles. Plugging these values into the ideal gas law equation, we can solve for T3:

T3 = (PiVf)/(nR) = (118 kPa)(668 cm^3)/(0.0045 mol)(8.3145 J/mol*K) = 7,969 K

Once we have the temperatures at each step, we can use the equation Q=nCv(delta T) to solve for Q for process 1 to 2. Here, we know the change in temperature (delta T) is T2-T1, and we can calculate the specific heat capacity (Cv) for a monoatomic gas using the equation Cv= (3/2)R, where R is the ideal gas constant.

Q(1-2) = (0.0045 mol)(1.5)(8.3145 J/mol*K)(2,619 K - 315 K) = 8,502 J

To find Ws for process 2 to 3, we can use the equation Ws = -nRTln(Vf/Vi). Here, we know the initial and final volumes (Vi and Vf) and the number of moles (n) from the given information. We can also use the ideal gas constant (R) and the temperature at step 2 (T2) that we calculated