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An LPG cylinder containing containing 15 kg butane at `27^(@)C` and 10 atm pressure is leaking. After one day, its pressure decreased to 8 atm. The quantity of the gas leaked isA. 1 kgB. 2 kgC. 3 kgD. 4 kg

Answer» Correct Answer - C
PV=nRT. Here, T and V are constant. Hence,
`(P_(1))/(P_(2))=(n_(1))/(n_(2))=(w_(1))/(w_(2))`.
`:. (10)/(8)=(15)/(w_(2))" or " w_(2)=12" kg"`.
Hence, gas leaked=15-12=3 kg


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