One mole of a gas is allowed to expand isothermally and reversibly from a volume of $1 d m^{3}$ to $50 d m^{3}$ at $273 K$ . Calculate W, $Δ E$ and q assuming the idea behaviour of the gas.

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Solution

Work done in an isothermal reversible process is given as-
$W = - 2.303 n R T log \frac{V_{2}}{V_{1}} . . . . . (1)$
Given:-
$n = 1 mole$
$T = 273 K$
$V_{1} = 1 {d m}^{3}$
$V_{2} = 50 {d m}^{3}$
$R = 8.314 J$
Substituting these values in ${e q}^{n} (1)$ , we have
$W = - 2.303 \times 1 \times 8.314 \times 273 \times log \frac{50}{1}$
$\Rightarrow W = - 5227.17 \times log 50 = 8.88 k J$
As the process is isothermal, i.e., temperature is constant.
$∴ Δ E = 0$
Now from first law of thermodynamics,
$Δ E = q + W$
$∵ Δ E = 0$
$\to q = - W = - (- 8.88) = 8.88 k J$
Hence the value of $W, Δ E$ and $q$ are $- 8.88 k J, 0$ and $8.88 k J$ respectively.

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