## 4C.3B

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### 4C.3B

Calculate the final temperature and the change in enthalpy when 765 J of energy is transferred as heat to 0.820 mol Kr(g) at 298 K and 1.00 atm (a) at constant pressure; (b) at constant volume. Treat the gas as ideal.

For part b, $\Delta$U = q, which is 765 J. Then, you use $\Delta$U to find $\Delta$H, which is
$\Delta$H = $\Delta$U + P$\Delta$V. Because it is at constant volume, shouldn't you just get $\Delta$H = $\Delta$U if $\Delta$V = 0? Why instead is the equation $\Delta$H = $\Delta$U +nR$\Delta$T?

Chem_Mod
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### Re: 4C.3B

The equation you mentioned only holds for changes in internal energy at constant pressure. At constant volume, you actually cannot calculate delta H because it is defined as the heat gained or lost at constant pressure. To calculate the final temperature, you would use the equation for temperature changes for an ideal gas at constant volume: q = n(Cv)delta T

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