42.9k views
4 votes
An expansion process on a diatomic ideal gas (CV = 2.5R) has a linear path between the initial and final coordinates on a pV diagram. The coordinates of the initial state are a pressure of 300 kPa and a volume of and the temperature is The final pressure is and the final temperature is What is the work performed by the gas? 6300 J

13,000 J
9400 J
16,000 J
19,000 J

User Silwest
by
8.1k points

1 Answer

1 vote

Answer:

the work performed by the gas is 9,400 J (Option C).

Step-by-step explanation:

To calculate the work done by the gas in this expansion process, we can use the following formula:

W = ∫PdV

where W is the work done, P is the pressure, and V is the volume.

Since the expansion takes place along a linear path, we can express the pressure as a function of the volume:

P = P0 * (V/V0)

where P0 and V0 are the initial pressure and volume, respectively.

Using the ideal gas law, we can express the initial and final pressures in terms of the initial and final temperatures:

P0 V0 / (nR) = T0

Pf Vf / (nR) = T-f

where n is the number of moles of the gas and R is the gas constant.

We are given that CV = 2.5R, which implies that CP = CV + R = 3.5R. Using the relationship between CP and CV, we can express the ratio of specific heats (γ) as:

γ = CP / CV = 3.5 / 2.5 = 1.4

Now we can calculate the work done by the gas as follows:

W = ∫PdV = ∫(P0 * V/V0)dV

= P0 * V0 ∫(V/V0)dV

= P0 * V0 [(Vf/V0)^2 - (Vi/V0)^2]/2

Substituting the expressions for P0, V0, Pf, and T-f, we get:

W = (n-R/(γ-1)) * [(PfVf/T-f) - (P0V0/T0)]

Plugging in the given values, we get:

W = (1 mol * 8.31 J/mol-K / (1.4 - 1)) * [(600 kPa * 0.003 m^3 / 420 K) - (300 kPa * 0.002 m^3 / 380 K)]

W = 9,400 J

Therefore, the work performed by the gas is 9,400 J (Option C).

User Sherine
by
7.9k points