Entropy of Ideal Gas N2 at 350 K, 2.0 bar

[mahdert]

Homework Statement

what is the entropy of 1.0 litre of gasoues n2 at Ta=350 K and Pa=2.0 bar given that So = 191.61 J/(K gmol) at To = 298.15 K and Po = 1 bar?

Homework Equations

I think : S(at 350) - So = NC_p ln (Ta/To) - NR ln(Pa/po)

The Attempt at a Solution

My problem is, C_p is not given and can we use this formula if p is not constant.

 

[Mapes]

1) Are you sure you don't know $c_P$ for an ideal gas?
2) To see if you can use this equation even if P is changing, how about starting from $dU=T\,dS-P\,dV$, isolating dS, and integrating?

 

[mahdert]

C_p = 5/2 R;

I did the integration and I got the right answer. But the question remains, the specific heat capacity at constant pressure; C_p should be used at constant pressure. Here, the pressure is changing? A qualitative expalnation would suffice. Thanks

 

[Mapes]

The pressure isn't changing; there's no process occurring. $c_P$ is just a variable.

 

[paranormal]

I can provide a response to this question by applying the principles of thermodynamics and statistical mechanics. The entropy of an ideal gas can be calculated using the following formula:

S = Nk ln(V/N) + Nk ln(T/T0) + Nk ln(U/NkT) + S0

where S is the entropy, N is the number of particles, k is the Boltzmann constant, V is the volume, T is the temperature, U is the internal energy, and S0 is the entropy at the reference state (in this case, To and Po).

In this case, we are given the temperature and pressure of the gas, but we also need to know the number of particles (N) and the internal energy (U) to calculate the entropy. The internal energy can be calculated using the ideal gas law:

U = NkT

We can also use the ideal gas law to find the number of particles:

PV = NkT

Therefore, N = PV/(kT).

Substituting these values into the entropy formula, we get:

S = (PV/(kT))k ln(V/(PV/(kT))) + (PV/(kT))k ln(T/T0) + (PV/(kT))k ln(PV/(kT)/NkT) + S0

Simplifying, we get:

S = PV/(kT) + Nk ln(T/T0) + Nk + S0

Now, we can plug in the given values for volume, temperature, pressure, and entropy at the reference state:

S = (1.0 L)(2.0 bar)/(1.38x10^-23 J/K)(350 K) + [(1.38x10^-23 J/K)(350 K)] ln(350 K/298.15 K) + [(1.38x10^-23 J/K)(350 K)] + 191.61 J/(K gmol)

S = 3.62x10^25 J/K + 0.30 J/K + 1.21x10^25 J/K + 191.61 J/(K gmol)

S = 1.21x10^25 J/K + 191.91 J/(K gmol)

Therefore, the entropy of 1.0 litre of gaseous N2 at Ta = 350 K and Pa = 2.0 bar