Calculating Delta S for a Monoatomic Gas System | Entropy Help

[stunner5000pt]

Calculate delta S for the system when the state of 3.00 mol of a monoatmoic perfect gas for which C_{p,m} = \frac{5}{2} R is changed from 298.15K and 1.00 atm to 398.15K and 5.00 atm.

Now \Delta S = n \ R ln \frac{V_{2}}{V_{1}} + n C_{v} ln \frac{T_{f}}{T_{i}}

but when i do that i get an answer that is -36 J/K which is not right the answer given 22.1 J/K.

what am i doing wrong i know that Cp is for constnat pressure but in this case the pressure is not constant! Neither is the volume. (at least it doesn't say otherwise)

 

[dextercioby]

Calculate delta S for the system when the state of 3.00 mol of a monoatmoic perfect gas for which C_{p,m} = \frac{5}{2} R is changed from 298.15K and 1.00 atm to 398.15K and 5.00 atm.

Now \Delta S = n \ R ln \frac{V_{2}}{V_{1}} + n C_{v} ln \frac{T_{f}}{T_{i}}

but when i do that i get an answer that is -36 J/K which is not right the answer given 22.1 J/K.

what am i doing wrong i know that Cp is for constnat pressure but in this case the pressure is not constant! Neither is the volume. (at least it doesn't say otherwise)

You're using the wrong formula.
1.Use the book from where u picked this formula to find the correct (for this case) formula.
2.If 1. is not an option,then u'll have to get is somehow using the formula which u already have (posted):
a) use Mandeleev-Clapeyron formula;
b) use Robert Mayer's law which gives the relation between the specific heats at constant pressure and constant volume and the gas constant R.
With these two,u can rewrite the formula already posted in terms of C_p and the ratio of pressures.

Yes,by the looks of it,it's a general process in which all 3 parameters vary.
Good luck!

 

[Andrew Mason]

Calculate delta S for the system when the state of 3.00 mol of a monoatmoic perfect gas for which C_{p,m} = \frac{5}{2} R is changed from 298.15K and 1.00 atm to 398.15K and 5.00 atm.

Your formula is correct. What are you using for V1 and V2 (or V2/V1)? The key here is to recognize that in this process, the volume changes as well. You have to determine the original and final volumes from the ideal gas law. Also C_v = 3/2 R The C_pm is a bit of a red herring.

AM

 

[stunner5000pt]

i was under the impression that Cp - Cv = nR where n is the number of moles of hte gas being used.

but that yields a negative heat capactiy which is wronggggg

i got the right answer using the formula i posted

i have two textbooks which i am using - One is the Fundamentals of Physics by HRW and one is Physical Chemistry by Atkins and De Paula but i think the physics textbook is serving me more in this chemistry course I am taking!

 

[Andrew Mason]

i was under the impression that Cp - Cv = nR where n is the number of moles of hte gas being used.

but that yields a negative heat capactiy which is wronggggg

No. That is the correct relation and it gives the right answer. Cp= 5/2 nR; Cv = 3/2 nR; Cp-Cv = 2/2 nR = nR; where Cp and Cv are the specific heats of the gas samples in J/K (not per mole). Again, what volumes (V2/V1) are you using?

AM

 

[stunner5000pt]

No. That is the correct relation and it gives the right answer. Cp= 5/2 nR; Cv = 3/2 nR; Cp-Cv = 2/2 nR = nR; where Cp and Cv are the specific heats of the gas samples in J/K (not per mole). Again, what volumes (V2/V1) are you using?

AM

i have got the right answer using the formula i was just confused about the CP Cv relation it's much clearer now thank you