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W = V 1 V 2 p d V = V 1 V 2 ( n R T V ) d V .
The figure is a plot of p on the vertical axis as a function of V on the horizontal axis. Two pressures are indicated on the vertical axis, p 1 and p 2, with p 1 greater than p 2. Two volumes are indicated on the horizontal axis, V 1 and V 2, with V 1 less than V 2.  Four points, A, B, C, and D are labeled. Point A is at V 1, p 1. Point B is at V 2, p 1. Point C is at V 2, p 2. Point D is at V 1, p 2. A straight horizontal line connects A to B, with an arrow pointing to the right indicating the direction from A to B. A straight vertical line connects B to C, with an arrow downward indicating the direction from B to C. A straight vertical line connects A to D, with an arrow pointing downward indicating the direction from A to D. A straight horizontal line connects D to C, with an arrow to the right indicating the direction from D to C. Finally, a curved line connects A to C with an arrow pointing in the direction from A to C.
The paths ABC , AC , and ADC represent three different quasi-static transitions between the equilibrium states A and C .

The expansion is isothermal, so T remains constant over the entire process. Since n and R are also constant, the only variable in the integrand is V , so the work done by an ideal gas in an isothermal process is

W = n R T V 1 V 2 d V V = n R T ln V 2 V 1 .

Notice that if V 2 > V 1 (expansion), W is positive, as expected.

The straight lines from A to B and then from B to C represent a different process. Here, a gas at a pressure p 1 first expands isobarically (constant pressure) and quasi-statically from V 1 to V 2 , after which it cools quasi-statically at the constant volume V 2 until its pressure drops to p 2 . From A to B , the pressure is constant at p , so the work over this part of the path is

W = V 1 V 2 p d V = p 1 V 1 V 2 d V = p 1 ( V 2 V 1 ) .

From B to C , there is no change in volume and therefore no work is done. The net work over the path ABC is then

W = p 1 ( V 2 V 1 ) + 0 = p 1 ( V 2 V 1 ) .

A comparison of the expressions for the work done by the gas in the two processes of [link] shows that they are quite different. This illustrates a very important property of thermodynamic work: It is path dependent . We cannot determine the work done by a system as it goes from one equilibrium state to another unless we know its thermodynamic path. Different values of the work are associated with different paths.

Isothermal expansion of a van der waals gas

Studies of a van der Waals gas require an adjustment to the ideal gas law that takes into consideration that gas molecules have a definite volume (see The Kinetic Theory of Gases ). One mole of a van der Waals gas has an equation of state

( p + a V 2 ) ( V b ) = R T ,

where a and b are two parameters for a specific gas. Suppose the gas expands isothermally and quasi-statically from volume V 1 to volume V 2 . How much work is done by the gas during the expansion?

Strategy

Because the equation of state is given, we can use [link] to express the pressure in terms of V and T . Furthermore, temperature T is a constant under the isothermal condition, so V becomes the only changing variable under the integral.

Solution

To evaluate this integral, we must express p as a function of V . From the given equation of state, the gas pressure is

p = R T V b a V 2 .

Because T is constant under the isothermal condition, the work done by 1 mol of a van der Waals gas in expanding from a volume V 1 to a volume V 2 is thus

W = V 1 V 2 ( R T V b a V 2 ) = | R T ln ( V b ) + a V | V 1 V 2 = R T ln ( V 2 b V 1 b ) + a ( 1 V 2 1 V 1 ) .

Significance

By taking into account the volume of molecules, the expression for work is much more complex. If, however, we set a = 0 and b = 0 , we see that the expression for work matches exactly the work done by an isothermal process for one mole of an ideal gas.

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Check Your Understanding How much work is done by the gas, as given in [link] , when it expands quasi-statically along the path ADC ?

p 2 ( V 2 V 1 )

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Internal energy

The internal energy     E int of a thermodynamic system is, by definition, the sum of the mechanical energies of all the molecules or entities in the system. If the kinetic and potential energies of molecule i are K i and U i , respectively, then the internal energy of the system is the average of the total mechanical energy of all the entities:

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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