Download Physical Chemistry I: Adiabatic Expansion of an Ideal Gas - Work Calculation - Prof. Marc and more Assignments Physical Chemistry in PDF only on Docsity!
The Richard Stockton College of New Jersey
Chemistry Program, School of Natural Sciences and Mathematics PO Box 195, Pomoma, NJ
CHEM 3410: Physical Chemistry I — Fall 2008
Homework 2 — Solutions
- For this problem we are dealing with an diatomic ideal gas, so Cp = 72 R and Cv = 52 R. The process would look something like this when the pressure is plotted versus the volume:
We can consider each step in the process separately.
1 to 1’ A good place to start would be to determine the number of moles of gas we are working with. We have T, P, and V of the initial state, so using the ideal gas law we can determine the number of moles of gas in our sample. Remember, we have to use R in the correct units. Since we are given the volume in cubic meters, it will be helpful to use Pa for pressure (1 atm = 101,325 Pa):
n =
P V
RT
(15 ∗ 101325 P a)(1 m^3 ) (8. 314 J/molK)(298)
= 615 moles
Now we can find P, V, and T for the final state (1’) using the information about the initial state. For an adiabatic expansion (thermally insulated) the relationship between the initial and final states is given by Equation 2.44:
PiV (^) iγ = Pf V (^) fγ
where γ = C Cpv = 75. Using the heat capacities for a diatomic ideal gas we get the final volume to be: V (^) fγ = PiV (^) iγ Pf
(15 ∗ 101325 P a)(1 m^3 )
(^75)
101325 P a Vf = 6. 92 m^3 With ideal gas law, we can find the final temperature, giving us the following values for P, V, and T:
Table 1: default 1 1’ P (Pa) 1. 52 × 106 1. 01 × 105 V (m^3 ) 1 6. T (K) 298 137
We can now find the work done in going from state 1 to 1’ in one of two ways. It is important to remember that the work is not a state function, so its value is path dependent.
Method 1: Use the definition of PV work. This involves a bit of integration, but isn’t too scary. We can write P as a function of V using Equation 2.44 as above:
P V γ^ = P 1 V 1 γ
P = P (V ) =
P 1 V 1 γ V γ Simple expansion expression work is given by:
w =
1
−P dV =
1
P 1 V 1 γ V γ^
dV = −P 1 V 1 γ
1
dV V γ
Performing the integration:
w =
[
P 1 V 1 γ γ − 1
V γ−^1
)] 1 ′
1
P 1 V 1 γ γ − 1
V 1 γ′−^1
V 1 γ−^1
Inserting values from the table above, we get:
w = − 2 .03 MJ
Method 2: It is an adiabatic expansion of an ideal gas, so q = 0. Therefore:
∆U = q + w = w
But for an ideal gas, the internal energy is only a function of temperature:
∆U = w =
nCv dT
Assuming Cv is constant over our temperature range, this reduces to:
w = nCv ∆T = nCv (T 1 ′ − T 1 )
Inserting out values from the table above, we get: w = − 2 .03 MJ Thankfully, the answers are identical. Both approaches are valid. The first is more general and would work in most cases. The second is a special case since we are dealing with an ideal gas.
1’ to 2 This step is done at constant pressure, so it is a bit more straight forward than above:
w =
1 ′
−P dV = −P
1 ′
dV = −P (V 2 − V 1 ′ )
To find V 2 we can use the ideal gas law again:
V =
nRT P
= 15 m^3
So, w = − 1. 01 × 105
15 m^3 − 6. 919 m^3
w = −819 kJ
The total work done in both processes is then:
−2030 kJ + −819 kJ = −2849 kJ
The sign of the work done in each step and the over all process should make sense. The gas is expanding, therefore it is doing work on the surroundings and work is “flowing” out of the system, hence the negative sign.
