Moderators: Chem_Mod, Chem_Admin

Posts: 50
Joined: Wed Nov 14, 2018 12:23 am


Postby kausalya_1k » Tue Feb 11, 2020 11:58 pm

Hi, I'm confused on question 4C.3:
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 vol- ume. Treat the gas as ideal.
More specifically, I was wondering, how would we go about solving part b's enthalpy? Also, is it just me, or is the solution manual incorrectly formatted for this question?

Harry Zhang 1B
Posts: 101
Joined: Sat Sep 14, 2019 12:16 am

Re: 4C.3

Postby Harry Zhang 1B » Thu Feb 13, 2020 11:13 pm

at constant pressure, delta H=q=765 J. The molar heat capacity of a monatomic ideal gas at constant pressure is 5/2 R, which is 20.785J/kmol. To calculate temperature change, use q/(C*the number of mols of Kr), which is equal to 44.89K. The final temp at constant pressure is therefore 343K. At constant volume , delta H=q+delta(PV). Work here is ignored because there is no work done without change in volume. The delta (PV) term can be converted to nRdeltaT by the ideal gas equation. The molar heat capacity at constant v is 1.5R, which is 12.471J/kmol. q/(C*the number of mols)=75K. q+nRdeltaT=1276=delta H. The final temp is 298+75=373K.

Daniel Yu 1E
Posts: 100
Joined: Sat Aug 24, 2019 12:15 am

Re: 4C.3

Postby Daniel Yu 1E » Sun Feb 16, 2020 10:21 pm

For part b, constant volume means that no work is done. Therefore Delta H = q+ nR*Delta T.

Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 2 guests