Question

Work done in reversible isothermal expansion for vander wall equation

Answer 1

Consider that one gram molecule of a perfect gas is taken in a cylinder having perfectly conducting walls and bottom, provided with a piston. Let the cylinder be placed on a source of heat at temperature T°A. If the piston is now moved slowly outwards, the gas expands, does some work and tends to cool but it absorbs required amount of heat from the source to keep it at the same temperature. The expansion is thus isothermal.

Let the piston of area of cross-section A move through a small distance dx so that the gas expands by a small amount dV. As the expansion is small, the pressure of the gas remains , practically, constant, say P.

Then the force acting on the piston, F = P×A

and small work done by the gas,

dW = F dx

= PA dx

dW = P dV [Since, A dx = dV]



P-V Diagram of Isothermal Expansion of an Ideal Gas

(a) Graphical Method
The isotherm PQ for the process is shown in below figure.

The net work done to expand the gas from a volume V1 to volume V2 will be,

W = Area ABDC

Thus, area occupied below the curve in between AB and CD gives the required value of work done.

(b) Analytical Method
Analytically, we can be calculated by integrating dW between the limits V1 to V2.

W=\int_{0}^{W}dW=\int_{v_1}^{v_2}PdV

For a perfect gas, PV = RT or P = RT/V

Substituting P = RT/V in the above equation, we get,

W = \int_{V^{_{1}}}^{V_{2}}\frac{RT}{V} dV

As temperature is constant during isothermal expansion and R is a gas constant,

W =RT \int_{V^{_{1}}}^{V_{2}}\frac{dV}{V} = RT [log_{e}V]_{V_{1}}^{V_{2}}

= RT (logeV2 -logeV1)

= RT loge (V2/V1)

W = 2.3026 RT log10 (V2/V1)

According to first law of thermodynamics, heat used up for the expansion of gas = W/J

So, H = 2.3026 (RT/J) log10 (V2/V1)

Since P1V1 = P2V2, thus, V2/V1= P1/P2

Substituting forV2/V1, H = 2.3026 (RT/J) log10 (P1/P2)

On the other hand, if the gas is compressed, the work is done on the gas and is therefore negative. The work done for isothermal compression is given by,

W = -2.3026 RT log10 (V2/V1)

So, W =2.3026 RT log10 (V1/V2)

Therefore, H = 2.3026 (RT/J) log10 (V1/V2)

and H = 2.3026 (RT/J) log10 (P2/P1)

Problem 1:-

Calculate the work done of n moles of a vander Waals gas in an isothermal expansion from Vi to Vf.

Concept:-

The work done W of gas in an isothermal expansion from volume Vi to Vf is defined as,

W = -∫ViVf p dV

The gas equation for van der Wall gas is,

(p+n2a/V2) (V-nb) = nRT

Here p is the pressure, V is the volume, R is the gas constant, T is the temperature, n is the number of moles, a and b are the van der Walls gas constant.

Solution:-

First we have to find out the pressure p of the gas.

From van der Walls gas equation (p+n2a/V2) (V-nb) = nRT, the pressure p will be,

p = (nRT/(V-nb)) – (n2a/V2)

To obtain the work done of n moles of a van der Walls gas in an isothermal expansion from volume Vi to Vf will be,

W = -∫ViVf p dV

= -∫ViVf [((nRT/(V-nb)) – (n2a/V2)] dV

= [- nRT ln(V-nb) – an2/V] ViVf (Since, ∫ (1/ V-nb) dV = ln(V-nb) and ∫ 1/V2 dV = - 1/V)

= (- nRT ln Vf –nb/ Vi –nb) –an2(1/Vf- 1/Vi)

From the above observation we conclude that, the work done of n moles of a van der Walls gas in an isothermal expansion from volume Vi to Vf would be (- nRT ln Vf –nb/ Vi –nb) –an2(1/Vf- 1/Vi).

Problem 2:-

A quantity of ideal gas occupies an initial volume V0at a pressure p0 and a temperature T0. It expands to volume V1 (a) at constant pressure, (b) at constant temperature, and (c) adiabatically. Graph each case on a pV diagram. In which case is Q greatest? In which case is W greatest? least? In which case is ?Eint greatest? Least?

Concept:-

Isobaric Process:-

The process in which, the change in volume and temperature of a gas takes place at a constant pressure is called an isobaric process.

Isothermal Process:-

The process in which, the change in pressure and volume takes place at constant temperature is called an isothermal change.

Adiabatic Process:-

The process in which, the change in pressure and volume and temperature takes place without any heat entering or leaving the system is called adiabatic change.

Solution:-

(a) Isobaric Process:-

The process in which, the change in volume and temperature of a gas takes place at a constant pressure is called an isobaric process.

A quantity of ideal gas occupies an initial volume V0 at a pressure p0 and a temperature T0. It expands to volume V1 at constant pressure. The pV diagram for this process as shown in below figure1.