# Isothermal Expansion Question

jetpackman

## Homework Statement

During an isothermal expansion, a confined ideal gas does -150J agaisnt its surroundings. Which of the following describes the heat transfer during this process?

A 150J of Heat was added to the gas
B 150J of Heat was removed from the gas
C 300J of Heat was added to the gas
D 300J of Heat was removed from the gas
E No heat

## Homework Equations

U = Q + W
In isothermal expansion, Q = -W

## The Attempt at a Solution

For this multiple choice problem, I thought the answer would be B. This is becaise the gas does -150J of work agaisnt its surroundings. This means that the surrounding did 150J of work into the gas. Thus, I would expect that if we Followed Q = -W, that 150J of heat was removed from the gas. However, the answer is (A).

physicsnoob93
Q must be positive. Because the gas does work against it's surroundings, the gas loses energy. So W is negative.

Q = -(-150) = 150

If you still cant see it
Q=-W
so W = -Q
-150 J = -Q
So Q = +150J.

Giving the answer A to be correct.

Hope this helped?

Staff Emeritus
Homework Helper
jetpackman, welcome to Physics Forums.

Something isn't right here. If the gas is expanding, it is doing a positive amount of work on the surroundings. Are you sure the problem statement says the gas is expanding?

At any rate, Q = -W as you said, since ΔU=0 for an ideal gas & isothermal process.

Homework Helper

## Homework Statement

During an isothermal expansion, a confined ideal gas does -150J agaisnt its surroundings. Which of the following describes the heat transfer during this process?

A 150J of Heat was added to the gas
B 150J of Heat was removed from the gas
C 300J of Heat was added to the gas
D 300J of Heat was removed from the gas
E No heat

## Homework Equations

U = Q + W
In isothermal expansion, Q = -W

## The Attempt at a Solution

For this multiple choice problem, I thought the answer would be B. This is becaise the gas does -150J of work agaisnt its surroundings. This means that the surrounding did 150J of work into the gas. Thus, I would expect that if we Followed Q = -W, that 150J of heat was removed from the gas. However, the answer is (A).
The convention is to write the first law as $\Delta Q = \Delta U + W$ where Q is the heat flow into the gas and W is the work done by the gas. U is the internal energy of the gas. In this case, there is no change in U, and there is an expansion of the gas so the gas does positive work on its surroundings. This means that the negative sign for work done by the gas on the surroundings is an error. The magnitude of the work is 150J.

$$\Delta Q = W = 150 J$$