Verified Solution • 2025–26 Session moderate

QJEEM20Pha47e: JEE Main 2020 Physics - Class 11 JEEM20Pha47e Solution: The change in the magnitude of the volume of an ideal gas when…

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

Step-by-Step Verified Solution

We know, PV=nRTPV = nRT

∴\therefore PΔV+VΔP=0P\Delta V + V\Delta P = 0 (for constant temp.)

and PΔVP\Delta V = nRΔTnR\Delta T (for constant pressure)

ΔT=PΔVnR\Delta T = {{P\Delta V} \over {nR}}

ΔP=−PΔVV\Delta P = - {{P\Delta V} \over V} ( ΔV\Delta V is same in both cases)

ΔTΔP=PΔVnRV−PΔV=−VnR=−TP{{\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 VnR=TP{{V \over {nR}} = {T \over P}} ]

∴\therefore ∣ΔTΔP∣=∣−3002∣=150\left| {{{\Delta T} \over {\Delta P}}} \right| = \left| {{{ - 300} \over 2}} \right| = 150

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