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
Answers
Answered by
267
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
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
Answered by
15
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.
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