4C.3B

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JamieVu_2C
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4C.3B

Postby JamieVu_2C » Sun Feb 02, 2020 9:26 pm

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, U = q, which is 765 J. Then, you use U to find H, which is
H = U + PV. Because it is at constant volume, shouldn't you just get H = U if V = 0? Why instead is the equation H = U +nRT?

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Re: 4C.3B

Postby Chem_Mod » Mon Feb 03, 2020 4:50 pm

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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