Monatomic gas, Isochoric Process.

AI Thread Summary
In an isochoric process involving 1.46 moles of a monatomic ideal gas, the work done on the system is zero due to constant volume conditions. The change in internal energy is equal to the heat added, as per the equation Change in I.E. = Q + W. The average kinetic energy formula, KE = (3/2)kT, is used to calculate the change in internal energy based on the temperature difference. The initial calculations mistakenly considered the change in kinetic energy for a single molecule instead of accounting for the total moles of gas. Correcting this will yield the accurate values for internal energy change and heat transfer.
yaylee
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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!
 
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(3/2)kT is the average translational KE of just one molecule.
 
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.
 
Thanks!
 
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