Fundamentals of thermodynamics

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Question

The work done in heating one mole of an ideal gas as constant pressure from $15^{\circ} C to 25^{\circ} C$ is

Moderate

Solution

$\begin{aligned} T_{1} = 15 + 273 = 288 K \\ T_{2} = 25 + 273 = 298 K \end{aligned}$
We know that $: P V_{1} = n R T_{1}$ and $P V_{2} = n R T_{2}$
Hence $V_{1} = \frac{n R T_{1}}{P}$ and $V_{2} = \frac{n R T_{2}}{P}$
$\begin{array}{l} ∴ V_{2} - V_{1} = \frac{n R T_{2}}{P} - \frac{n R T_{1}}{P} i.e., \\ V_{2} - V_{1} = \frac{n R}{P} (T_{2} - T_{1}) \end{array}$
Or $Δ V = \frac{n R}{P} (T_{2} - T_{1})$
But $W = - P Δ V = - P \times \frac{n R}{n} (T_{2} - T_{1})$
$= - n R (T_{n} - T_{1})$
$\begin{aligned} = - n R (T_{2} - I_{1}) \\ W = - 1 mol \times 1.987 {calK}^{-} {mol}^{-} \\ \times (298 - 288) K = 19.87 cal \end{aligned}$
So, the correct answer is (d)

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