1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Todd Minehardt Nov 2 '16 at 21:04
3
$\begingroup$

You did everything correctly with the exception of converting the units of your result for work.

First, calculate the work in joules:

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

Second, add the heat absorbed (already in joules):

$$\Delta E = 200\,\mathrm{J} - 1823.85\,\mathrm{J} = -1623.85\,\mathrm{J}$$

Third, always remember to check your units!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.