Home
Class 11
CHEMISTRY
An LPG cylinder, containing 15 kg butanc...

An LPG cylinder, containing 15 kg butance at `27^@C` and 10 atm pressure, is leaking. After one day, its pressure decreased to 8 atm. The quantity of gas leaked is

A

1 kg

B

2 kg

C

3 kg

D

4 kg

Text Solution

Verified by Experts

The correct Answer is:
C

According to ideal gas equation, `PV = nRT`
`:. (P_(1))/(P_(2)) = (n_(1))/(n_(2))` [where V, T are constant]
or `(P_(1))/(P_(2)) = (w_(1))/(w_(2))`
Hence `(10)/(8) = (15)/(w_(2)) " or " w_(2) = 12 kg`
`:.` Gas leaked `= 15 - 12 = 3 kg`
Promotional Banner

Similar Questions

Explore conceptually related problems

An L.P.G cylinder contains 15kg of butane gas at 27^(@)C and 10 atm pressure It was leaking and its pressure fell down to 8 atm pressure after one day Calculate the amount of leaked gas .

At 27^(@)C and 8.0 atm pressure, the density of propene gas is :

At 27^(@)C and 3.0 atm pressure, the density of propene gas is :

At 27^(@)C and 4.0 atm pressure, the density of propene gas is :

At 27^(@)C and 6.0 atm pressure, the density of propene gas is :

At 27^(@)C and 2.0 atm pressure, the density of propene gas is :

At 27^(@)C and 5.0 atm pressure, the density of propene gas is :

At 27^(@)C and 7.0 atm pressure, the density of propene gas is :

Calculate the density of SO_2 " at " 27^@C and 1.5 atm pressure.

A 1 mol gas occupies 2.4L volume at 27^(@) C and 10 atm pressure then it show :-