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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?


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.


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 ) .


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:

Questions & Answers

What is differential form of Gauss's law?
Rohit Reply
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
Abu Reply
i have to say. who cares. lol. why know that t all
is this just a chat app about the openstax book?
Lord Reply
kya ye b.sc ka hai agar haa to konsa part
MPL Reply
what is charge quantization
Mayowa Reply
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
is the time quantized ? how ?
What do you meanby the statement,"Is the time quantized"
Can you give an explanation.
there are some comment on the time -quantized..
time is integer of the planck time, discrete..
planck time is travel in planck lenght of light..
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
determine absolute zero
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
Opeyemi Reply
U can easily calculate work done by 2.303log(v2/v1)
Amount of heat added through q=ncv^delta t
Change in internal energy through q=Q-w
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
it is the work done in moving a charge to a point from infinity against electric field
Ashok Reply
what is the weight of the earth in space
peterpaul Reply
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Thats why it can't have a constant value of g ....
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
please why is the first law of thermodynamics greater than the second
Ifeoma Reply
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
Mateshwar Reply
explain the lack of symmetry in the field of the parallel capacitor
Phoebe Reply
pls. explain the lack of symmetry in the field of the parallel capacitor
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