JEE Main 1-Sep-2021 Shift 2 Solved Paper

Weight of empty LPG cylinder = 14.8 kg
Weight of full LPG cylinder = 29 kg
∴ Weight of gas = 29 − 14.8 = 14.2 kg
If weight of full LPG cylinder = 23 kg
then weight of gas used = 29 − 23 = 6 kg at ambient temperature.
From ideal gas equaiton, pV = nRT
or pV=

Weight‌of‌solute

Molecular‌mass‌of‌solute

×RT
orpV=

W

M

×RT
Applying ideal gas to LPG cylinder when gas is full,
pV=nRT
3.47‌atm×V=

14.2‌kg

M

×RT ...(i)
Applying ideal gas to LPG cylinder when gas is reduced to 23 kg at ambient temperature,
pV=nRT
p×V=

8.2‌kg

M

×RT ...(ii)
Divide Eq. (i) by (ii)

3.47×V

p×V

=

14.2‌kg‌RT M

8.2‌kg×RT M

3.47

p

=

14.2

8.2

⇒ p=

3.47×41

71

=2.003‌atm
Hence, answer is 2.