# Solving for Heat flow in Reversible Isobaric Expansion

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.

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