For an ideal gas the density of the gas is ρ₀ when temperature and pressure of the gas are T₀ and P₀ respectively. When the temperature of the gas is 2T₀, its pressure will be 3P₀. The new density will be (a) 3/2ρ₀ (b) 4/3ρ₀ (c) 3/4ρ₀ (d) 2/3ρ₀

Answer: (a). ρ = PM/RT ⇒ ρ ∝ P/T. (ρ)/(ρ₀) = (3P₀)/(P₀)·(T₀)/(2T₀) = 3/2. So the answer is (a) 3/2ρ₀.

For an ideal gas the density of the gas is ρ₀ when temperature and… | MHT CET 2023 (13 May, Shift 1)

Q48MHT CET 2023 · 13 MAY · SHIFT 1Medium

Gas Laws & Ideal Gas Equation

For an ideal gas the density of the gas is $\rho_0$ when temperature and pressure of the gas are $T_0$ and $P_0$ respectively. When the temperature of the gas is $2T_0$ , its pressure will be $3P_0$ . The new density will be

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Answer: (a)

$\rho = \frac{PM}{RT} \Rightarrow \rho \propto \frac{P}{T}$ .
$\frac{\rho}{\rho_0} = \frac{3P_0}{P_0}\cdot\frac{T_0}{2T_0} = \frac{3}{2}$ .
So the answer is (a) $\frac{3}{2}\rho_0$ .

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Gas Laws & Ideal Gas Equation question from MHT CET 2023 (13 May, Shift 1)

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