# Thermaldynamics problem, reversible process work done on

## Homework Statement

One mole of a monatomic ideal gas is taken through a reversible process from an initial state "a" to a final state "b" during which the pressure of the gas varies with volume V as shown below. Calculate the work Won[/on](a->b) done on the gas during this process.

## Homework Equations

problem shows a graph which i'll just write out points a and b

a = (300kPa, 1m^3), b= (500kPa, 2m^3)

## The Attempt at a Solution

work on = negative integral Pextdv and since this is reversible then P should equal Pext

I can calculate the temp at a and b, but i don't think that will contribute to anything.

Sorry this should be easy, but i'm confused

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Andrew Mason
Homework Helper

## Homework Statement

One mole of a monatomic ideal gas is taken through a reversible process from an initial state "a" to a final state "b" during which the pressure of the gas varies with volume V as shown below. Calculate the work Won[/on](a->b) done on the gas during this process.

## Homework Equations

problem shows a graph which i'll just write out points a and b

a = (300kPa, 1m^3), b= (500kPa, 2m^3)

## The Attempt at a Solution

work on = negative integral Pextdv and since this is reversible then P should equal Pext

I can calculate the temp at a and b, but i don't think that will contribute to anything.

Sorry this should be easy, but i'm confused

Is it a straight line from a to b or is it a curve? If it is a curve, we need to know the relationship between P and V at all points from a to b.

AM

sorry it's a straight line, and i thought to myself, if there is no heat added or lost, then could i use the fact

dU = Q + W where Q = 0

then (3/2)nRdT = dU = W?

Just to point out, $\Delta$E = Q - W... not Q + W.
From introductory general physics, $\Delta$U = -W.
Potential energy is equal to negative work. $\Delta$U = Ua - Ub or $\Delta$U = -(Ub - Ua)
W = Ub - Ua

Thus, $\Delta$U = -W.

If I'm not mistaken, correct?

Andrew Mason
Homework Helper
sorry it's a straight line, and i thought to myself, if there is no heat added or lost, then could i use the fact

dU = Q + W where Q = 0

then (3/2)nRdT = dU = W?
How does it increase its volume AND pressure (hence temperature) without positive heatflow into the gas? Is the gas doing positive work on the surroundings or is positive work being done on the gas?

AM

Andrew Mason
Homework Helper
Just to point out, $\Delta$E = Q - W... not Q + W.
From introductory general physics, $\Delta$U = -W.
Potential energy is equal to negative work. $\Delta$U = Ua - Ub or $\Delta$U = -(Ub - Ua)
W = Ub - Ua

Thus, $\Delta$U = -W.

If I'm not mistaken, correct?
This only applies if the process is adiabatic, which this is not. Whether there is a - sign for W depends on the convention used. There are two different conventions for the first law. You are using the convention I prefer, which is that dW = PdV = work done BY the gas. Some texts use dW = -PdV = work done ON the gas. The OP is using the latter convention. The question actually asks for the work done ON the gas ie $-\int PdV$.

AM

This only applies if the process is adiabatic, which this is not. Whether there is a - sign for W depends on the convention used. There are two different conventions for the first law. You are using the convention I prefer, which is that dW = PdV = work done BY the gas. Some texts use dW = -PdV = work done ON the gas. The OP is using the latter convention. The question actually asks for the work done ON the gas ie $-\int PdV$.

AM
Oh, right. Sign conventions always trips me up.
Thanks for clarifying, and sorry to Liquidxlax for adding confusion :P

Oh, right. Sign conventions always trips me up.
Thanks for clarifying, and sorry to Liquidxlax for adding confusion :P
no problem i have no problem with the sign convention, the fact i just made W positive wasn't for a fact i was confused about the convention.

thanks for the help guys

Andrew Mason