Question

A 2.24L cylinder of oxygen at N.T.P. is found to develop a leakage. When the leakage was plugged the pressure dropped to 570mm of Hg. The number of moles of gas that escaped will be

Answer 1

Original pressure of Hg= 760
drop in pressure= 570
change in pressure= 760-570= 190mm
= 190/760 atm

PV= nRT
n = P V/ RT
= 190 x 2.24/ 760 x 0.821 x 273 
= 0.0249 moles 

Answer 2

Answer:

0.025 mol

Explanation:

Given :

Volume V = 2.24 L

Pressure dropped = 570 mm of Hg

We have :

Original pressure P  = 760 mm of Hg

Change in pressure Δ P = 760 - 570 = 190 mm of Hg

From ideal gas equation :

P V = n R T

= > n =  Δ P / P .  V / R T

= > n = ( 190 / 760  × 2.24 ) / ( 0.0821 × 273 ) mol

= > n = 425.6 / 17034.108 mol

= > n = 0.024985 mol

= > n = 0.025 mol

Therefore , number of moles of gas that escaped will be 0.025 mol.