Name: Factors Affecting the Internal Energy
Uploaded: 2019-07-19
Duration: 3 min 47 s
Description: The internal energy of a thermodynamic system is the energy contained within it. Let's watch the video to learn more about the internal energy of a system.

Question

Derive the expression for the maximum work.

Accepted Answer

Consider $n$ moles of an ideal gas enclosed in an ideal cylinder fitted with a massless and frictionless movable rigid piston. Let $V$ be the volume of the gas at a pressure $P$ and a temperature $T$ .

If in an infinitesimal change pressure changes from $P$ to $P - dP$ and volume increases from $V$ to $V + dV$ . Then the work obtained is, $dW = - (P - dP) dV = - PdV + dPdV$ .

Since $dP . dV$ is negligibly small relative to $PdV$ .

$dW = - PdV$ .

Let the state of the system change from $A (P_{1}, L_{1})$ to $B (P_{2}, V_{2})$ isothermally and reversibly, at temperature $T$ involving number of infinitesimal steps.

$\underset{P_{1}, V_{1}}{A} \overset{T}{\to} \underset{P_{2}, V_{2}}{B}$ .

Then the total work or maximum work in the process is obtained by integrating above equation.

$W_{\max} = \int_{A}^{B} dW . W_{\max} = \int_{A}^{B} - PdV [∵ PV = nRT] . ∴ P = \frac{nRT}{V} . W_{\max} = \int_{V_{1}}^{V_{2}} (- nRT) \frac{dV}{V} . W_{\max} = (- nRT) \int_{V_{1}}^{V_{2}} \frac{dV}{V} . W_{\max} = (- nRT) ({InV}_{2} - {InV}_{1}) . W_{\max} = (- nRT) \log_{e} (\frac{V_{2}}{V_{1}}) . ∴ W_{\max} = - 2.303 {nRTlog}_{10} (\frac{V_{2}}{V_{1}}) .$

At constant temperature,

$∵ P_{1} \times V_{1} = P_{2} \times V_{2} . ∵ \frac{P_{1}}{P_{2}} = \frac{V_{2}}{V_{1}} . W_{\max} = - 2.303 {nRTlog}_{10} (\frac{P_{1}}{P_{2}}) .$

Where $n, P, V$ and $T$ represents number of moles, pressure, volume and temperature respectively. For the process,

$ΔU = 0, ΔH = 0$ .

The heat absorbed in reversible manner $Q_{rev}$ , is completely converted into work.

$Q_{rev} = - W_{\max}$ .

Hence, work obtained is maximum.

Derive the expression for the maximum work.

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