Answer the following question. Derive the expression for the maximum work. - Chemistry

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Answer in Brief

Answer the following question.

Derive the expression for the maximum work.

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Solution

1. Consider n moles of an ideal gas enclosed in a cylinder fitted with a frictionless movable rigid piston. It expands isothermally and reversibly from the initial volume V1 to final volume V2 at temperature T. The expansion takes place in a number of steps as shown in the figure.

2. When the volume of a gas increases by an infinitesimal amount dV in a single step, the small quantity of work done

dW = -Pext dV        ....(1)

3. As the expansion is reversible, P is greater by a very small quantity dp than Pext.

Thus P - Pext = dP or Pext = P - dP     ....(2)

Combining equations (1) and (2),

dW = - (P - dP)dV = - PdV + dP.dV

Neglecting the product dP.dV which is very small, we get

dW = - PdV         .....(3)

4. The total amount of work done during the entire expansion from volume V1 to V2 would be the sum of the infinitesimal contributions of all the steps. The total work is obtained by integration of Equation (3) between the limits of initial and final states. This is the maximum work, the expansion being reversible.

Thus,

\[\int\limits_{\text{initial}}^{\text{final}}\text{dW} = - \int\limits_{\text{V}_1}^{\text{V}_2} \text{PdV}\]

Hence,

Wmax = - \[\int\limits_{\text{V}_1}^{\text{V}_2} \text{PdV}\]      .....(4)

5. Using the ideal gas law, PV = nRT,

Wmax = - \[\int\limits_{\text{V}_1}^{\text{V}_2} \text{nRT} \frac{\text{dV}}{\text{V}}\]

= -\[\text{nRT} \int\limits_{\text{V}_1}^{\text{V}_2} \frac{\text{dV}}{\text{V}}\]     ...(∵ T is constant.)

= - nRT ln `("V")_("V"_1)^("V"_2)`

= - nRT (ln V2 - ln V1)

= - nRt ln `"V"_2/"V"_1`

= - 2.303 nRT log10 `"V"_2/"V"_1`     ....(5)

6. At constant temperature, P1V1 = P2V2  or `"V"_2/"V"_1 = "P"_1/"P"_2`

Replacing `"V"_2/"V"_1` in equation (5) by `"P"_1/"P"_2`, we get,

Wmax = - 2.303 nRT log `"P"_1/"P"_2`     ....(6)

Equations (5) and (6) are expressions for work done in reversible isothermal process. 

Concept: Concept of Maximum Work
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Chapter 4: Chemical Thermodynamics - Exercises [Page 88]

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Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercises | Q 4.01 | Page 88
SCERT Maharashtra Question Bank 12th Standard HSC Chemistry Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Short answer questions (Type- II) | Q 2.2
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