$10$ mole of an ideal gas at $27^{o} C$ & $10 a t m$ pressed occupying a volume $24.6 l$ undergoes the following changes.
a) isothermal a reverable expansion to $264 l$
b) isothermal a irreversable expansion to $246 l$
c) isochoric heating to $177^{o} C$
Calculate the work done in each of these process.

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Solution

(i) Work done in isothermal reversible expansion= $- 2.303 n R T {log}_{10} \frac{V_{2}}{V_{1}}$
$= - 2.303 \times 10 R \times 300 {log}_{10} \frac{246}{24.6}$
$\approx - 57 K j$
(ii) Work done in isothermal irreversible expansion=
$- P Δ V$
$- 10 \times (246 - 24.6)$
$- 221.4 J$
(iii)Work done for isochoric pressure= $0$
as the $W = - P Δ V$ , $Δ V = 0$ in this case.

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10 mole of an ideal gas at 27oC & 10 atm pressed occupying a volume 24.6l undergoes the following changes.a) isothermal a reverable expansion to 264l b) isothermal a irreversable expansion to 246lc) isochoric heating to 177oCCalculate the work done in each of these process.

(i) Work done in isothermal reversible expansion= −2.303nRTlog10V2V1=−2.303×10R×300log1024624.6≈−57Kj(ii) Work done in isothermal irreversible expansion=−PΔV−10×(246−24.6)−221.4J(iii)Work done for isochoric pressure=0as the W=−PΔV, ΔV=0 in this case.