Monatomic gas, Isochoric Process.

[yaylee]

Homework Statement

n = 1.46 moles of ideal gas are heated isochorically (at constant volume) from tepmerature To = 649 oC to temperature Tf = 1184 oC. Find:

a) Work done on the system.
b) Change in Internal Energy of the system.
c) The total heat, Q, added or removed from the system.

Homework Equations

Change in I.E. = Q + W
KE (avg) = T = 3/2kT, where k = 1.38 x 10^-26 kJ/K

The Attempt at a Solution

a) Since this is an isochoric system, work done is ZERO! (No issues here.)
b,c) Change in Internal Energy, therefore is equal to Q, by the first equation, Change in IE = Q + W.
So, we can calculate change in IE, by change in KE.

KE = (3/2kT) for a monatomic gas.
so, change in KE = 3/2(k)(Tfinal - Tinitial), = (3/2)(1.38E-26)(1500 K - 704K) = change in IE = Q = 1.65 x 10^-23.

Am I reading the problem incorrectly? Many thanks in advance once again!

 

[TSny]

(3/2)kT is the average translational KE of just one molecule.

 

[SteamKing]

It appears you have calculated the change in KE for a single molecule of the gas.
Remember, you have 1.46 moles of gas to start with.

 

[yaylee]

Thanks!

 

[Nickie]

Your approach and calculations are correct. Since this is an isochoric process, there is no change in volume and therefore no work is done. The change in internal energy is equal to the heat added or removed from the system, and this can be calculated using the average kinetic energy formula for a monatomic gas. Your final answer of 1.65 x 10^-23 kJ is the change in internal energy, and it is also the total heat added or removed from the system. Well done!