Seventh Edition ? 4C.3

Moderators: Chem_Mod, Chem_Admin

Ashley Bouma 1F
Posts: 30
Joined: Fri Sep 28, 2018 12:21 am

Seventh Edition ? 4C.3

Postby Ashley Bouma 1F » Sat Feb 02, 2019 2:57 pm

Calculate the final temp and the change in enthalpy when 765J of energy is transferred as heat to .820 mol Kr(g) at 298K and 1.00 atm at constant pressure. How should I approach this problem?

Xuan Kuang 2L
Posts: 31
Joined: Wed Nov 14, 2018 12:23 am

Re: Seventh Edition ? 4C.3

Postby Xuan Kuang 2L » Sun Feb 03, 2019 1:27 pm

To approach this problem, you'd first have to recognize that you're using the formula qp=nCp,mdeltaT. You would then substitute given values, with 765 J being qp, 0.820 mol for n, and 298 K for initial temperature. However, in this problem, you'd also have to recognize the rotational contribution of Kr because it is treated as an ideal gas, which is this case would be 5/2R (you can view an explanation on this in 4C.3 I believe). Thus, then set-up for this equation would be 765 J =(0.820 mol)(5/2 x 8.314 J mol-1 K-1)(Tfinal-298K), and you should get Tfinal= 373 K.

Also at constant pressure, we know that change in enthalpy is equal to qp, so we say that deltaH= qp= 765 J

Hope this helps!

Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest