A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air.
A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air. As it rises in the air external pressure decreases and the volume of the gas increases till finally it bursts when external pressure is 0.3bar. What is the limit at which volume of the balloon can stay inflated ?
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Given :
P1 = Initial pressure = 1 bar
V1 = Initial volume = 2.27 L
P2 = Final pressure = 0.3 bar
To find :
V2 = Final volume
Formula :
P1V1 = P2V2 (at constant n and T)
Calculation :
According to Boyle’s law,
P1V1 = P2V2 (at constant n and T)
∴ V2 = \(\frac{P_1V_1}{P_2}\)
= \(\frac{1\times 2.27}{0.3}\)
= 7.566667 L ≈ 7.567 L
∴ The balloon can stay inflated below the volume of 7.567 L.
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