# Calculating the work done in a PV diagram

• jisbon
In summary, the conversation discusses the problem of obtaining work done in AB and the volume at B in an adiabatic process, when the temperature at B is unknown. The formula for an adiabatic process, pV^γ=constant, is used to solve for the constant and ultimately the volume at B. The conversation also mentions using the formula for change in internal energy to find the temperature at B, and concludes with a reminder to use PV=nRT to find the temperature at B.
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?

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 :)
kuruman said:
For an adiabatic process ##pV^{\gamma}=constant##. Can you find the constant?

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

jisbon said:
Is this correct/is there a shorter way? Thanks
Can you get T at B from PV = nRT?

TSny said:
Can you get T at B from PV = nRT?
oh gosh. Guess I was just too blured :/ Thanks :)

## What is a PV diagram?

A PV diagram, also known as a pressure-volume diagram, is a graphical representation of the relationship between pressure and volume in a thermodynamic system. It is commonly used in the analysis of work done by the system.

## How is work calculated in a PV diagram?

In a PV diagram, the work done is equal to the area under the curve. This means that to calculate work, you must find the area of the shape formed by the curve on the diagram.

## What are the units of work in a PV diagram?

The units of work in a PV diagram are Joules (J) or Newton-meters (Nm). This is because work is calculated by multiplying the force (measured in Newtons) by the displacement (measured in meters).

## Can negative work be done in a PV diagram?

Yes, negative work can be done in a PV diagram. This occurs when the volume decreases while the pressure increases, resulting in negative area under the curve. This can happen, for example, when a gas is compressed.

## What factors affect the work done in a PV diagram?

The work done in a PV diagram is affected by the change in pressure and volume of the system. A larger area under the curve indicates a greater amount of work done, while a smaller area indicates less work done. The type of process, such as isothermal or adiabatic, also affects the work done.

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