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Introduction-thermodynamics

Question

The change in the magnitude of the volume of an ideal gas when a small additional pressure $Δ P$ is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $Δ T$ at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm. respectively. If $|Δ T| = C |Δ P|$ then value of C in (K/atm.) is

Difficult

Solution

$\begin{array}{l} At constant temperature PV = nRT \\ P Δ V + V Δ P = 0 \\ \Rightarrow Δ V = - V \frac{Δ P}{P} - - - - (1) \\ At constant pressure \frac{V}{T} = \frac{nR}{P} \Rightarrow V \propto T \\ \Rightarrow \frac{ΔV}{V} = \frac{Δ T}{T} \\ \Rightarrow Δ V = V \cdot \frac{Δ T}{T} - - - - (2) \\ F r o m e q u a t i o n (1) a n d e q u a t i o n (2) \\ ∵ V \frac{Δ T}{T} = - V \frac{Δ P}{P} \\ \Rightarrow |\frac{ΔT}{ΔP}| = \frac{T}{P} = \frac{300}{2} = 150 \end{array}$

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