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

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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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Fundamentals of thermodynamics

Remember concepts with our Masterclasses.

The work done in heating one mole of an ideal gas as constant pressure from15∘C to 25∘C is

T1=15+273=288KT2=25+273=298KWe know that :PV1=nRT1 and PV2=nRT2Hence V1=nRT1P and V2=nRT2P ∴ V2−V1=nRT2P−nRT1P i.e., V2−V1=nRPT2−T1Or ΔV=nRPT2−T1But W=−PΔV=−P×nRnT2−T1 =−nRTn−T1 =−nRT2−I1 W=−1mol×1.987calK−mol−×(298−288)K=19.87calSo, the correct answer is (d)

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