To determine variables in adiabatic reversible process do we need to look up heat capacity?

  • #1
zenterix
629
82
Homework Statement
I have a quick question about a relatively simple problem.

Suppose we have 1 mole of a specified gas (say, helium) and it undergoes a reversible adiabatic process.

We are given initial volume ##V_1## and final volume ##V_2## and initial temperature ##T_1##.
Relevant Equations
If we want to calculate ##T_2## and ##P_2##, is is mandatory that we determine what ##\gamma=\frac{C_P}{C_V}## is?
This problem does not specify if helium is to be treated as an ideal gas. I am assuming it is an ideal gas.

We can determine ##P_1## from the ideal gas law

$$P_1=\frac{nRT_1}{V_1}\tag{1}$$

Since the process is adiabatic we have ##\delta q=0## and

##\dU=C_V dT=-PdV=-\frac{nRT}{V}dV\tag{2}##

After integrating and some algebra we reach

$$\frac{T_2}{T_1}=\left (\frac{V_1}{V_2}\right )^{\gamma-1}\tag{3}$$

From this we can sub in for temperature using the ideal gas law to reach

$$\left (\frac{V_1}{V_2}\right )^\gamma=\frac{P_2}{P_1}\tag{4}$$

$$PV^\gamma=\text{constant}=k\tag{5}$$

We don't know what ##k## because we don't know what ##\gamma is##.

If we knew gamma we would have ##k## from ##P_1V_1^\gamma## and we could then compute ##P_2##.

We could also then compute ##T_2## from (3).

To find ##\gamma## we can do

$$\gamma=\frac{C_P}{C_V}=\frac{C_P}{C_P-nR}$$

and so we just need to look up ##C_P## for helium at ##T_1##.

In the integration to reach (3) we already assumed that ##C_V## is constant which means ##C_P=C_V+nR## must be constant.

This is how I solved the problem.

I just want to make sure there isn't another way. Do we need to look up heat capacity to solve this problem in this way or is there another way?
 
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  • #2
Cp and Cv are heat capacities per mole, so nCvdT=~PdV and Cp-Cv=R. Also, He is monoatomic, so Cv=1.5R
 

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