Question

4). (a). Calculate an expression for the work done during an isothermal, reversible expansion for a gas which is described using the van der Waals equation of state.

(b). The van der Waals constants for a gas are a = 506.5 kPa L2 mol2 and b = 6.0x10−2 L mol−1. Determine the work done by 2.0 moles of a gas that expands from 1.5 L to 10 L at 325 K.

(c). The a constant is attributed to attractive forces in the gas while the b constant is attributed to the repulsive forces in the gas. For the same conditions as before, if a were doubled, what is the amount of work done by the gas? Explain why this result makes sense.

Answer 1

Ans (a) Starting from the definition of work,

As the process is reversible, therefore Pext = P

(1)

Using Van der Waals gas equation -:

Solving this equation for P, we get

Substituting this value of P in equation of work (1), we get:

Assume that the expansion of gas takes place from V1 to V2, thus put limits in the integration:

Because the process is isothermal, T is constant. Thus it can be taken out along with other constants:

Integrating leads to the result:

Sol (b) -:

Converting pressure in Pa to bar, we get -:

b = 0.06 Lmol-1

Thus use

V1 = 1.5 L and V2 = 10 L

T = 325 K

n = 2

Substitute values we get:

w = - 7.5116 - 11.46 = 18.971 J

(c) If a is doubled, that means a now is 2 * 5.06 = 10.12

Then w = -7.5116 - 22.92 = 30.4316 J

means magnitude of work is increased when a is doubled ( Here negative sign represents that work is done by the system because gas is expanding)

Work increases because of increase in attractive forces in the gas. More are the attractive forces, more will be the collisions between gas molecules leading to more expansion and thus more work.