# Change in internal energy of a gas upon expansion against a constant external pressure

I have a question about the following problem which I got wrong:

During expansion of its volume from $\mathrm{1.00\,L}$ to $\mathrm{10.00\,L}$ against a constant external pressure of $\mathrm{2.00\,atm}$, a gas absorbs $\mathrm{200\,J}$ of energy as heat. Calculate the change in internal energy of the gas.

First, I used the equation: change in $E = q + w$. I calculated work ($-P \Delta V$) and found it to be $$-18 = -2 \cdot (10 - 1)$$. Then, I added $-18 + q$, the heat absorbed ($\mathrm{200\,J}$). This gave me a result of $\mathrm{-182\,J}$, which is wrong by a large factor. My textbook gives the answer of $\mathrm{-1623\,J}$, but I am not sure where I am going wrong...

Do you know where I made a mistake? Thank you!

• My initial guess is units: atm probably need to be converted to Pa. 2 atm = 202.65 kPa and 1623/202.65 = 8, which makes me suspicious (you almost never get integers!). I don't have time to work it through right now but check your units. – Todd Minehardt Nov 2 '16 at 21:04

$$w = -p\Delta V = (-2\,\mathrm{atm}) \times 9\,\mathrm{L} = \left(-18\,\mathrm{L\cdot atm}\right) \left({101.325\,\mathrm{J}\over\mathrm{L\cdot atm}}\right) = -1823.85\,\mathrm{J}$$
$$\Delta E = 200\,\mathrm{J} - 1823.85\,\mathrm{J} = -1623.85\,\mathrm{J}$$