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

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Heat & Thermodynamics

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 ....

Answer: 150

For an ideal gas

P V = n R T

. At constant temperature

P d V + V d P = 0

gives

∣Δ V ∣ = \frac{V}{P} ∣Δ P ∣

, while at constant pressure

P d V = n R d T

gives

∣Δ V ∣ = \frac{n R}{P} ∣Δ T ∣

. Equating the two volume changes,

\frac{V}{P} ∣Δ P ∣ = \frac{n R}{P} ∣Δ T ∣

, and substituting

V = \frac{n R T}{P}

leaves

∣Δ T ∣ = \frac{T}{P} ∣Δ P ∣

. Thus

C = \frac{T}{P} = \frac{300}{2} = 150 K at m^{- 1}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Integer answer