A balloon contains 500 m ³ of He at 27° C and 1 atmospheric pressure. Then, volume of He at -3° C and 0.5 atmospheric pressure will be (a) 700 m ³ (b) 900 m ³ (c) 1000 m ³ (d) 500 m ³

Answer: (b). Gas equation: (P₁V₁)/(T₁) = (P₂V₂)/(T₂), with T₁ = 300 K, T₂ = 270 K. V₂ = V₁ · (P₁)/(P₂) · (T₂)/(T₁) = 500 × 1/0.5 × (270)/(300) V₂ = 500 × 2 × 0.9 = 900 m ³ So the answer is (b) 900 m ³.

A balloon contains 500 m ³ of He at 27° C and 1 atmospheric pressure… | MHT CET 2026 (13 May, Shift 2)

Q2MHT CET 2026 · 13 MAY · SHIFT 2Easy

Gas Laws & Ideal Gas Equation

A balloon contains $500$ m $^{3}$ of He at $27^\circ$ C and $1$ atmospheric pressure. Then, volume of He at $-3^\circ$ C and $0.5$ atmospheric pressure will be

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Answer: (b)

Gas equation: $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ , with $T_1 = 300$ K, $T_2 = 270$ K.
$V_2 = V_1 \cdot \dfrac{P_1}{P_2} \cdot \dfrac{T_2}{T_1} = 500 \times \dfrac{1}{0.5} \times \dfrac{270}{300}$
$V_2 = 500 \times 2 \times 0.9 = 900$ m $^3$
So the answer is (b) $900$ m $^3$ .

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Gas Laws & Ideal Gas Equation question from MHT CET 2026 (13 May, Shift 2)

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