## Quiz 1 Prep Question 3

Kendall_Chaffin_3C
Posts: 25
Joined: Wed Sep 21, 2016 2:59 pm

### Quiz 1 Prep Question 3

In Quiz 1 Prep Question 3 it asks: If 2.00 mol of an ideal gas at 300 K and 3.00 atm expands isothermally and reversibly from 6.00 L to 18.00 L and has a final pressure of 1.20 atm, what is w,q, and change in internal energy?

I found w using w=-nRTln(V2/V1) but how do I find q without a change in temperature? Or does q=0?

Also, is there anywhere I can find solutions to the quiz prep in the back of the course reader?

EllisJang2O
Posts: 18
Joined: Wed Sep 21, 2016 2:55 pm

### Re: Quiz 1 Prep Question 3

Because this is an isothermal reversible reaction, temperature remains constant: ΔT = 0. In an isothermal process (if the gas is ideal), internal energy does not change because the amount of energy entering the surroundings is equal to the amount entering the system. Therefore ΔE = 0.
So from the equation ΔE = q + w, if ΔE = 0, then q = -w.

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