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Solve the following. A balloon is inflated with helium gas at room temperature of 250 C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in the air. - Chemistry

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प्रश्न

Solve the following.

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 the 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 the volume of the balloon can stay inflated?

योग
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उत्तर

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 = `("P"_1"V"_1)/"P"_2=(1xx2.27)/0.3` = 7.566667 L ≈ 7.567 L

The balloon can stay inflated below the volume of 7.567 L.

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अध्याय 10: States of Matter - Exercises [पृष्ठ १५९]

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बालभारती Chemistry [English] Standard 11 Maharashtra State Board
अध्याय 10 States of Matter
Exercises | Q 5. (A) | पृष्ठ १५९

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