Question 620: Thermodynamics & KTG

Question

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 _________.

Accepted Answer

We know, $$PV = nRT$$
$$ 	herefore $$ $$P\Delta V + V\Delta P = 0$$ (for constant temp.)
and $$P\Delta V$$ = $$nR\Delta T$$ (for constant pressure)
$$\Delta T = {{P\Delta V} \over {nR}}$$
$$\Delta P =  - {{P\Delta V} \over V}$$ ($$\Delta V$$ is same in both cases)
$${{\Delta T} \over {\Delta P}} = {{P\Delta V} \over {nR}}{V \over { - P\Delta V}} = {{ - V} \over {nR}} =  - {T \over P}$$
[As PV = nRT
$$ \Rightarrow $$ $$ {{V \over {nR}} = {T \over P}} $$]
$$ 	herefore $$ $$\left| {{{\Delta T} \over {\Delta P}}} ight| = \left| {{{ - 300} \over 2}} ight| = 150$$

Thermodynamics & KTG

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