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 _________. Numerical answer type. Work it out, then open the answer.

Answer: 150. Constant T: V = nRT/P ⇒ |Δ V| = (nRT)/(P²)|Δ P|. Constant P: |Δ V| = nR/P|Δ T|. Equate: |Δ T| = T/P|Δ P| ⇒ C = T/P = 300/2 So C = 150 K/atm.

The change in the magnitude of the volume of an ideal gas when a… | JEE Main 2020 (4 Sep, Shift 2)

Q67JEE Main 2020 · 4 SEP · SHIFT 2Difficult

Gas Laws & Ideal Gas Equation

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 _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 150

Constant $T$ : $V = \frac{nRT}{P} \Rightarrow |\Delta V| = \frac{nRT}{P^2}|\Delta P|$ .
Constant $P$ : $|\Delta V| = \frac{nR}{P}|\Delta T|$ .
Equate: $|\Delta T| = \frac{T}{P}|\Delta P| \Rightarrow C = \frac{T}{P} = \frac{300}{2}$
So $C = 150$ K/atm.

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Gas Laws & Ideal Gas Equation question from JEE Main 2020 (4 Sep, Shift 2)

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