# Finding work of adiabatic compressor using ideal gas

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1. Jan 29, 2016

### worryingchem

1. The problem statement, all variables and given/known data
Methane at $P_1$ and $T_1$ is compressed to a pressure of $P_2$ adiabatically at steady-state. Calculate the work done on the compressor and the temperature $T_2$ of the discharge gas. Use ideal gas model.
Given:
$T_1, P_1, P_2, C_p, \gamma = 1.4$

$PV^\gamma = constant$

$W = \int_{P_1}^{P_2} VdP$

$\Delta H = \int_{T_1}^{T_2} C_p dT = W$

2. Relevant equations
$dH = TdS + VdP$

$dQ = TdS$

Steady-state energy balance: $\Delta H = W_{shaft}$

3. The attempt at a solution
Hello,
my first question is that is this process reversible?
Because if I started from $dH = TdS + VdP$, in order to apply $0 = dQ = TdS$ to get $dH = VdP$, won't the process have to be reversible?

Then, using the energy balance, I make $dH = VdP = dW$ and got the given equation $W = \int_{P_1}^{P_2} VdP$. My second question here is that can we use $dW = -PdV$ here instead of $dW = VdP$? And why can't or can we? I'm confused as to when we can use $dW = -PdV$.

2. Jan 29, 2016

### Staff: Mentor

You're supposed to assume the process is reversible.
Yes.

You are supposed to be calculating the shaft work, not the total work which, if you remember, also includes PV work pushing the gas into the compressor and it pushing gas ahead of it out of the compressor. For this situation, the equation gives you the shaft work.

But there is also another way to do this problem that gives the same answer for the shaft work. This is where you treat each parcel of gas passing through the compressor as a closed system experiencing an adiabatic reversible compression. On this basis, you can calculate the exit temperature from the compressor (as they ask you to do). Then you calculate ΔH from $mC_p ΔT = W_s$. When you solve it this way, you should get the same answer as when you integrate VdP.

3. Jan 29, 2016

### worryingchem

If I treat the parcel of gas as a closed system, will the energy balance be like the following?
$dU = -PdV$
$C_v \Delta T = -\int_{P_1}^{P_2} PdV$
From here, will I plug in $PV^{\gamma} = constant$ to find $T_2$ ?

4. Jan 29, 2016

### Staff: Mentor

You write $PV^{\gamma} = constant$ and you substitute $V=RT/P$

5. Jan 30, 2016

### worryingchem

So I get $P_2(RT_2/P_2)^\gamma = constant$ and solve for $T_2$.
Thank you for clearing this up for me.