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Standard XII

Chemistry

Ideal Gas Equation

An LPG cyli...

Question

An $L P G$ cylinder, containing $15 k g$ butane at $27^{\circ} C$ and $10 a t m$ pressure, is leaking. After one day, its pressure decreased to $8 a t m$ . The quantity of gas leaked is :

1 k g

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2 k g

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3 k g

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4 k g

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Solution

The correct option is C $3 k g$
We know that the ideal gas equation :

$P V = n R T$

Where, P = pressure
V = volume of gas
n = moles
R = Ideal gas constant
T = temperature

$n = \frac{m}{M}$

So,
$\frac{P}{m} = K$

$P_{1} m_{2} = P_{2} m_{1}$

$10 \times m_{2} = 8 \times 15$

$m_{2} = 12 k g$

So, the amount of gas leaked $= 15 - 12 = 3 k g$ .

Hence, the correct option is C.

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An LPG cylinder, 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 gas leaked is :

The correct option is C 3 kgWe know that the ideal gas equation :PV=nRTWhere, P = pressure V = volume of gas n = moles R = Ideal gas constant T = temperature n=mMSo,Pm=KP1m2=P2m110×m2=8×15m2=12 kgSo, the amount of gas leaked =15−12=3 kg.Hence, the correct option is C.

An $L P G$ cylinder, containing $15 k g$ butane at $27^{\circ} C$ and $10 a t m$ pressure, is leaking. After one day, its pressure decreased to $8 a t m$ . The quantity of gas leaked is :