help...MOLAR ENTROPY- PLEASE HELP AS SOON AS POSSIBLE.. URGENT?

Question

The molar entropy of helium gas at 25°C and 1.00 atm is 126.1 J K-1 mol-1. Assuming ideal behavior, calculate the entropy of the following.

(a) 0.178 mol He(g) at 25°C and a volume of 4.61 L

(b) 2.37 mol He(g) at 25°C and a volume of 4219.3 L

Answer 1

Do you understand and be able to do work in calculus?

(a) Use the ideal gas formula. 0.178 mol He at 25°C and 1.00 atm occupies:
V = nRT/P = 0.178*0.08206*298 = 4.35 (L).
For 0.178 mol He at 25°C to have a volume of 4.61 L, its pressure must be;
P = nRT/V = 0.178*0.08206*298/4.61 = 0.944 (atm)
We are given the molar entropy of helium at 25°C and 1.00 atm: 126.1 J/K mol. The gas needs to expand from 4.35L to 4.61L. The work done is:
[integral](4.35 to 4.61) PdV
= [integral](4.35 to 4.61) nRTdV/V
= nRT ln(4.61/4.35)
= 0.178*8.314*298 ln(4.61/4.35)
= 25.6 J
Since the internal energy of the gas does not change at constant T, the heat must be provided by the environment to the system at 25°C or 298K, enable the system to do the work of 25.6J. Hence based on the formula:
(New S) = (Old S) + (delta S) = (Old S) + Q/T
the entropy of the gas is:
(0.178 mol)*(126.1 J/K mol) + (25.6 J)/298K
= 22.5 J/K

Please practice (b) by yourself as done for (a).