Calculating the work done in a PV diagram

[jisbon]

Homework Statement

2 moles of diatomic ideal gas goes through a cyclic process as shown below. From A to B (non-standard) , B to C (adiabatic), C to A (isothermal). Calculate work done by the gas in the process AB.

Relevant Equations

pV=nRT

So I've been digging this problem for quite some time, and still can't figure out a way to obtain work done in AB.
I do understand that work done in AB = area under the graph. However, I can't figure out how to obtain the volume at B. I can't use PV=nRT since I am unable to obtain the temperature at B. I am however able to obtain the temperature at A and C at 607.7K since it is a isothermal process. I tried summing change in internal energy =0, but can't proceed due to AB being a unstandard equation. What can I do in this case?

 

[kuruman]

For an adiabatic process $pV^{\gamma}=constant$. Can you find the constant?

 

[jisbon]

Solved the problem as 176J. Thanks for the reminder for this formula :)

For an adiabatic process $pV^{\gamma}=constant$. Can you find the constant?

 

[jisbon]

EDIT: An additional question, they want me to find the temperature at B. Since an adiabatic equation has changed in internal energy = -work is done, I used the following formula but seemed to get the correct answer.

$$\begin{aligned}\dfrac {5}{2}nR\Delta t=-W\\ =\dfrac {1}{1-\gamma }\left( P_{f}V_{t}-P_{i}V_{i}\right) \end{aligned} $$

Assuming x is the temperature at B, and volume at B found in part (a) to be 0.06095, and temperature at A to be 607.7K,

$$\dfrac {5}{2}\left( z\right) \left( 8.31\right) \left( 607.7-x\right) =\dfrac {1}{1-1.4}(\left( 0\cdot 1\right) \left( 1\right) \left( 1.01\times 10^{5}\right) -\left( 2\right) \left( 0.00095\right) \left( 1.01\times 10^{5}\right)) $$

Is this correct/is there a shorter way? Thanks

 

[TSny]

Is this correct/is there a shorter way? Thanks

Can you get T at B from PV = nRT?

 

[jisbon]

Can you get T at B from PV = nRT?

oh gosh. Guess I was just too blured :/ Thanks :)