An ideal gas in a cylinder was placed in a heater and gained 5.50 kJ of energy as heat. If the cylinder increased in volume from 345 mL to 1846 mL against an atmospheric pressure of 750. Torr during this process, what is the change in internal energy of the gas in the cylinder?
I am unsure of which equation to use for this problem. Can anyone explain the process? Thanks!
Textbook 4B.5
Moderators: Chem_Mod, Chem_Admin
-
ashna kumar 3k
- Posts: 103
- Joined: Fri Sep 24, 2021 5:34 am
- Been upvoted: 1 time
Re: Textbook 4B.5
For this question we use the w=-Pext*(delta V) equation. We know to use this because there is a change in volume with a constant external pressure, 750 Torr.
To solve we do:
w=-(750 Torr* 1atm/760Torr)((1846-345)x10^-3))=-1.48 L.atm
To convert this to Joules we use the ideal gas law constants:
(-1.48 L.atm)(8.134 J.K.mol-1/0.08206 L.atm.K-1)=-1.50 x 10^2J= -0.150 kj.
Thus the total change in internal energy is 5.50 kj (q)- 0.150 kj= 5.35 kj.
To solve we do:
w=-(750 Torr* 1atm/760Torr)((1846-345)x10^-3))=-1.48 L.atm
To convert this to Joules we use the ideal gas law constants:
(-1.48 L.atm)(8.134 J.K.mol-1/0.08206 L.atm.K-1)=-1.50 x 10^2J= -0.150 kj.
Thus the total change in internal energy is 5.50 kj (q)- 0.150 kj= 5.35 kj.
Return to “Calculating Work of Expansion”
Who is online
Users browsing this forum: No registered users and 5 guests