-1
$\begingroup$

This question was given on my Practice midterm:

Calculate $q,w,\Delta U$ and $\Delta H$ for a reversile isobaric expansion from (1.00 Bar, 20.0 L) to (1.00 bar, 40.0 L)

Equations used:

$q_p=C_p\Delta T$ $; C_{p Ideal} = \frac {5}{2}R$ where $C_p$ is heat capacity per mol. so $\frac {C_{p Ideal}}{n} = \frac {5}{2}R$

I solved the problem like this:

$w=-P\Delta V = -2000J$ [correct answer] $$q_p=C_p\Delta T$$ $$\Delta T = \Delta [\frac {PV}{nR}]=\frac {1}{nR}P\Delta V$$
$$q_p=C_p\Delta T= \frac {5Rn}{2}\frac {1}{nR}P\Delta V= 5/2*P\Delta V = 5/2*2000J = 5000J$$ Which is the wrong answer. The teacher's work goes like this. I dont know where she got n=2mol

$$T_1=\frac {P_1V_1}{nR}=\frac{10^5*20*10^{-3}}{2*8.314}=120.3K$$ $$T_2=\frac {P_2V_2}{nR}=\frac{10^5*40*10^{-3}}{2*8.314}=240.6K$$ $$q_p=C_p\Delta T = (2.00 mol)*3.5(8.314)(120.3K)= 7.00kJ$$

I was originally thinking that she might have made a typo and wrote 3.5 for 5/2 instead lf 2.5, but she magically found n = 2 mols.

Please tell me how to get the correct answer.

Thanks in advance.

$\endgroup$
1
$\begingroup$

As far as the 3.5 is concerned, $C_v=\frac{5}{2}R$, $C_p-C_v=R$, so $C_p=\frac{7}{2}R$.

Regarding the 2 moles, were you told either the number of moles or the initial temperature?

$\endgroup$
  • $\begingroup$ You're completely correct. I did not have the entire question. The other information was written on a different PDF. I'm sorry for wasting time !! $\endgroup$ – Jess L Oct 18 '16 at 0:06

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.