Question:
A balloon contains \(1500\ \mathrm{m}^3\) of helium at \(27^{\circ}\mathrm{C}\) and 4 atmospheric pressure. The volume of helium at \(-3^{\circ}\mathrm{C}\) temperature and 2 atmospheric pressure will be
A balloon contains \(1500\ \mathrm{m}^3\) of helium at \(27^{\circ}\mathrm{C}\) and 4 atmospheric pressure. The volume of helium at \(-3^{\circ}\mathrm{C}\) temperature and 2 atmospheric pressure will be
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Always convert temperature to Kelvin: \(T(K) = T(^{\circ}C) + 273\).
Updated On: Apr 23, 2026
- \(2700\ \mathrm{m}^3\)
- \(1900\ \mathrm{m}^3\)
- \(1700\ \mathrm{m}^3\)
- \(1500\ \mathrm{m}^3\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Use ideal gas equation \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\).
Step 2: Detailed Explanation:
Given: \(V_1 = 1500\ \text{m}^3\), \(T_1 = 27 + 273 = 300\ \text{K}\), \(P_1 = 4\ \text{atm}\), \(T_2 = -3 + 273 = 270\ \text{K}\), \(P_2 = 2\ \text{atm}\).
\(V_2 = \frac{P_1V_1}{T_1} \times \frac{T_2}{P_2} = \frac{4 \times 1500}{300} \times \frac{270}{2} = 20 \times 135 = 2700\ \text{m}^3\).
Step 3: Final Answer:
Thus, \(V_2 = 2700\ \text{m}^3\).
Use ideal gas equation \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\).
Step 2: Detailed Explanation:
Given: \(V_1 = 1500\ \text{m}^3\), \(T_1 = 27 + 273 = 300\ \text{K}\), \(P_1 = 4\ \text{atm}\), \(T_2 = -3 + 273 = 270\ \text{K}\), \(P_2 = 2\ \text{atm}\).
\(V_2 = \frac{P_1V_1}{T_1} \times \frac{T_2}{P_2} = \frac{4 \times 1500}{300} \times \frac{270}{2} = 20 \times 135 = 2700\ \text{m}^3\).
Step 3: Final Answer:
Thus, \(V_2 = 2700\ \text{m}^3\).
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