The change in the magnitude of the volume of an ideal gas when a small additional pressure $\Delta$P is
applied at a constant temperature, is the same as the change when the temperature is reduced by
a small quantity $\Delta$T at constant pressure. The initial temperature and pressure of the gas were 300
K and 2 atm. respectively.
If |$\Delta$T| = C|$\Delta$P| then value of C in (K/atm.) is _________.
Answer (integer)
150
Solution
We know, $PV = nRT$<br><br>$\therefore$ $P\Delta V + V\Delta P = 0$ (for constant temp.)<br><br>and $P\Delta V$ = $nR\Delta T$ (for constant pressure)<br><br>$\Delta T = {{P\Delta V} \over {nR}}$<br><br>$\Delta P = - {{P\Delta V} \over V}$ ($\Delta V$ is same in both cases)<br><br>$${{\Delta T} \over {\Delta P}} = {{P\Delta V} \over {nR}}{V \over { - P\Delta V}} = {{ - V} \over {nR}} = - {T \over P}$$<br><br>[As PV = nRT<br><br>$\Rightarrow$ ${{V \over {nR}} = {T \over P}}$]<br><br>$\therefore$ $$\left| {{{\Delta T} \over {\Delta P}}} \right| = \left| {{{ - 300} \over 2}} \right| = 150$$
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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