- #1

duds1234

- 3

- 0

## Homework Statement

A gas is compressed at a constant pressure of 0.639 atm from 5.41 L to 2.3 L. In the process, 470 J of energy leaves the gas by heat.

a) What is the work done on the gas (J)?

b) What is the change in its internal energy(J)?

## Homework Equations

a) ##W=-PΔV##

b) ##ΔU=Q+W##

## The Attempt at a Solution

a)

##P = 0.639atm {\frac{1Pa}{9.872*10^-6atm}} = 64728.52512Pa##

##ΔV = V_f - V_i = 2.3L - 5.41L = -3.11L {\frac{1*10^{-3}m^3}{1L}} = -0.00311m^3##

##W = -P ΔV = -6.4728.52512Pa * -0.00311m^3 = 201.3057131J##

I submitted my answer to part a and it was correct. Part b is what I need help with. I believe that I am doing it correctly, but my answer is returned as incorrect. My incorrect work and answer appears below.

b)

##ΔU = Q + W = -407J + 201.3057131J = -205.6942869J##

Any insight into why my work/answer for part b is incorrect would be greatly appreciated!