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### Pressure – Volume Work:

- Consider an ideal gas having definite mass (say n moles) be enclosed in a cylinder fitted with weightless, frictionless, tightly fitted, movable piston. Let ‘A’ be the area of the cross-section of the cylinder.
- Let the gas expand from volume V
_{1}to V_{2}against constant external pressure Pext which is exerted on the piston. Due to expansion, the piston moves upward by a distance ‘d ‘ . In this process, the system loses energy to the surroundings. Hence the work done by the system is negative.

Work done = – Opposing Force x displacement

∴ W = – F x displacement ……. (1)

Now, pressure is force per unit area.

∴ Pressure = Force / Area

∴Force = Pressure x Area

∴ F = P x A …….. (2)

substituting this value in equation (1)

Work = – Pressure x Area x displacement

∴ W = – P x A x d ………. (3)

But , A x d = Volume through gas expands = Change in volume = ( V_{2} – V_{1}) = Δ V

Substituting this value in equation (3)

W = – P_{ext} ΔV

- This is an expression for work done in terms of pressure and volume in the isothermal expansion of an ideal gas against constant external pressure.

#### Notes:

- It is the external pressure against gas expands i.e. P
_{ext}and not the pressure of the gas that is used in evaluating the work done. - The final equation shows that work W depends only on the change in volume (ΔV) and opposing pressure (P
_{ext}) and not on the quantity of the gas (i.e. it is independent of the number of moles) and Temperature of the gas (T). - The work done by the system in a process depends on the way or manner in which it is carried out. Work, therefore, is not a state function. It is path function.

### Free Expansion:

- The free expansion is also known as work of expansion in a vacuum. When a gas expands in a vacuum there is no opposing force i.e. P
_{ext}= 0. - For getting work done opposing force is necessary, hence no work is done when a gas expands in a vacuum. such an expansion is called as free expansion.

W = – P_{extΔ}V

∴ W = 0 × ΔV = 0

### Work Done in a Cyclic Process:

- A cyclic process is one which consists of a series of intermediate steps, at the end of which the system returns to its initial state.
- Since in a cycle, initial and final states are the same.

∴ Δ(state function) = 0

∴ ΔU = 0 (i.e. No change in internal energy)

Now, ΔU = q + W

∴ 0 = q + W

∴ q = -W

- Thus in a cyclic process heat absorbed by the system from the surrounding (q) is equal to work done (W) by the system on the surrounding.

### Concept of Maximum Work:

- According to the first law of thermodynamics, ΔU = q + W In an isothermal process, ΔU = 0, ∴ q = – W
- Therefore, all the heat absorbed by the system is utilized to do work. (i.e. maximum utilization of the energy takes place. )
- Work of expansion is given by the product of external pressure and the volume change. W = – P
_{ext}ΔV - In any expansion, the external pressure must be less than the pressure of the gas. If the external pressure is zero, the work done is also zero as the gas expands into the vacuum. If the external pressure is increased gradually, more and more work will be done by the gas during expansion. If the external pressure becomes equal to the pressure of the gas, there will be no change in the volume and thus ΔV = 0.Work done is also zero. If P
_{ext}is more than the pressure of the gas the gas cannot expand. Therefore, when P_{ext}becomes P then ΔV will be maximum. In reversible process, P_{ext}is always less than the pressure of the gas, by an infinitesimally small quantity.

W = (P – dp ) dV

- In the equation W tends to the maximum as (P – dp) tends to P or dp tends to zero. Therefore work done in an isothermal reversible expansion of an ideal gas is maximum work.

#### Conditions for Maximum Work:

- All the changes taking place in a system during the process are reversible. All the changes taking place in a system during the process should take place in infinitesimally small infinite steps. During the change, the driving and opposing forces should defer by infinitesimally small amount. The system remains in mechanical equilibrium with the surroundings.

#### Expression for Maximum Work:

- Work done in an isothermal reversible expansion is maximum work.

- Consider ‘n ‘ moles of an ideal gas enclosed in a cylinder fitted with a weightless, frictionless, air tight movable piston. Let the pressure of the gas be P which is equal to external atmospheric pressure P. Let the external pressure be reduced by an infinitesimally small amount dp and the corresponding small increase in volume be dV. so the small work done in the expansion process

dW = – Pext. dV

∴ dW = – (P – dp). dV

∴ dW = – (P.dV – dp.dV) ….. (1)

Since both dp and dV are very small, the product dp.dV is very small and can be neglected in comparison with P.dV.

Then above equation becomes dW = – P.dV ………. (2)

- When the expansion of the gas is carried out reversibly then there will be series of such p.dV terms. The total maximum work W
_{max}can be obtained by integrating this equation between the limits V_{1}to V_{2}. Where V_{1}is initial volume and V_{2}is final volume.

For an isothermal expansion , Boyle’s law is applicable.

Hence P_{1}V_{1} = P_{2}V_{2} i.e. V_{2} / V_{1} = P_{1} / P_{2}

Where P_{1} and P_{2} are initial and final pressure respectively.

#### Notes:

- As the ratio of volumes or pressure is used in the above equation when work done is to be calculated the volumes and the pressures may be expressed in any unit, provided both the quantities are expressed in the same unit.
- The work obtained using the above equation will be in joule if R is taken in S.I. unit.
- The number of moles = Wt. in gram / Molecular wt.in grams
- Absolute temperature T k = to C + 273

#### Difference Between Heat and Work:

- When heat is added to gas the random motion of molecules of gas increases, thus heat is a stimulus to gas which increases the random motion of the gas molecules. Thus heat is a random form of energy.
- Let us consider work done on a system by compressing the gas in a cylinder. The movement of the piston makes the gas molecules to move in the direction of applied force. Thus work is stimulus which increases the organized motion of the molecules of the gas. Thus work is organized form of energy.

#### Work is Path Function:

- The work done in isothermal constant pressure process is given by W = – Pext ΔV i.e. W = – Pext ( V
_{2}– V_{1}) Where, Pext = External opposing pressure V_{1}= Initial volume V_{2}= Final volume. - The work done in the isothermal reversible process is given by

- Where, n = Number of moles of the gas R = Universal gas constant T = Absolute temperature of the gas V
_{1}= Initial volume V_{2}= Final volume - We can see that the work done depends on the manner or the conditions under which the process carried out. Thus it is not dependent on initial and final conditions of the system but on the path followed by the system. Hence work is path function or it is not a state function.

#### Sign Convention Used in Thermodynamics:

- During the expansion, V
_{2}> V_{1}, Hence ΔV is positive. Hence work done W is negative.Thus when the work is done by the system on the surrounding i.e. Work of expansion is taken as negative (- W). - During compression V
_{2}< V_{1}, Hence ΔV is negative.Hence work done W is positive.Thus when the work done on the system by the surrounding i.e. Work of compression is taken as positive (+W).

Science > Chemistry > Chemical Thermodynamics and Energetics > You are Here |

good practice

Thanks for your comment. We are working to help as many students as we can. I would like to receive suggestions and corrections from your side. If you like the article share it in your group using social media. Let us make quality education free. Be the part of our movement.

thanks for the work done…

I need help here,

how to derive an expression for the heat absorbed during a reversible isothermal expansion of an ideal gas ?

Thank you MUGUME BENARD for your comment. On the same page proof of expression for maximum work is given. It is the same as work done in the isothermal reversible process. Still, have any doubt? you can ask it anytime.

Please I have one question…

One mole of an ideal gas is allowed to expand reversibly, i.e., against a confining pressure that is at all times infinitesimally less than the gas pressure 0.1 atm . The temperature being kept constant at 273k… How much work is done by the gas

Sir external pressure is nothing but atmospheric pressure or what ???? Can we decrease atmospheric pressure?? And how ?

We can not decrease atmospheric pressure. Generally, In chemistry, we carry reactions at atmospheric pressure. Hence external pressure is atmospheric pressure. If the reaction is carried out in special enclosures where we can change the pressure, and in such case, the external pressure can be altered. Reactions carried out in a vacuum is one such example, where external pressure is zero.

why we neglect integration constant at the derivation of maximum work?

When we are deriving formula for maximum work we are using limits V1 and V2 for the integration. So this is a case of definite integration. In definite integration in last step we deduct final value of the function from initial value and thus the constant of integration is eliminated.

Thanks sir.

Sir from where the value 2.303 come?

∫dx/x = logx + c here it is log is natural log to the base e, while we use log table which is defined to base 10. To convert the natural log to the base e into log to base 10 we multiply the quantity by 2.303.

Please give the formula of irreversible?

W = P(V2 – V1) = n1RT = n2RT = (n1 – n2)RT

In deriving the equation of max work done in isothermal reversible expansion ,can you use p external to equal to final p .having in mind p final is larger infinitesimally

A reversible process does not mean final pressure and initial pressure differ by infinitesimal value but it means that the pressure is changed by infinitesimal value dp in continuous steps from initial pressure P1 to final pressure P2. It is a process of integration of dp in the limit P1 to P2.

when calculating using the derived formula -nRT ln v2/v1, how can I convert my final answer to joules and calories.

The unit of the final answer depends on the unit of the universal gas constant R. If its unit is J K-1 mol-1, the answer obtained is in joules. If its unit is Kcal K-1 mol-1, the answer obtained is in Kcal. If its unit is cal K-1 mol-1, the answer obtained is in calories.

Conversion:

joule ÷ 4.188 = calories

calories x 4.188 = joules

Thank you for the clarification it means a lot to me