JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 3)
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 _________.
If |$$\Delta $$T| = C|$$\Delta $$P| then value of C in (K/atm.) is _________.
Answer
150
Explanation
We know, $$PV = nRT$$
$$ \therefore $$ $$P\Delta V + V\Delta P = 0$$ (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
$$ \Rightarrow $$ $$ {{V \over {nR}} = {T \over P}} $$]
$$ \therefore $$ $$\left| {{{\Delta T} \over {\Delta P}}} \right| = \left| {{{ - 300} \over 2}} \right| = 150$$
$$ \therefore $$ $$P\Delta V + V\Delta P = 0$$ (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
$$ \Rightarrow $$ $$ {{V \over {nR}} = {T \over P}} $$]
$$ \therefore $$ $$\left| {{{\Delta T} \over {\Delta P}}} \right| = \left| {{{ - 300} \over 2}} \right| = 150$$
Comments (0)
