Due to leakage in a 5 litre container pressure changed from 3.0 to 2.5 atm. Initially CH
gas filled in this container at 25°C. How much CH, in g has been leaked during the
period? Also report the volume leaked if collected at 1 atm and 25°C
Answers
Answer:
The amount of methane gas leaked is 1.6352 grams.
The volume of gas leaked is 2.5 litres.
Explanation:
Given data:
There is a leakage in a 5 litre container.
Thus the pressure changes from 3 atm to 2.5 atm.
Initial temperature of methane gas = 25°C
= 25+273 = 298 K
R = 0.0821
To find:
1) amount of methane gas in grams that has been leaked during the
period
2) the volume leaked if collected at 1 atm and 25°C.
Solution :
Consider case 1
Pressure = 3atm
We know the formula
PV = nRT
3*5 = n1 * 0. 0821 * 298
15 = n1 * 24.4658
n1 = 15/24.4658
n1 = 0.6131
Similarly let us consider case 2
Pressure = 2.5tm
We know the formula
PV = nRT
2.5*5 = n2 * 0. 0821 * 298
12.5 = n2 * 24.4658
n2 = 12.5/24.4658
n2= 0.5109
Difference in moles =n2-n1
= 0.6131 -0.5109
= 0.1022 moles
Now we have to find out amount of methane gas leaked
We know the formula
Moles = mass/molecular mass
Molecular mass of CH4 that is methane gas = 16
Thus according to the formula,
0.1022 = mass/16
Mass of methane gas leaked = 0.1022*16
= 1.6352 grams.
Now we need to find the volume of gas leaked
Pressure = 1 atm
We know the ideal gas equation
PV = nRT
V = nRT/P
= 0.1022*0.0821*298/1
=2.5004/1
= 2.5 L
The volume of gas leaked is 2.5 L.
Hence,
The amount of methane gas leaked is 1.6352 grams.
The volume of gas leaked is 2.5 litres.
Solution :
Pressure = 3 atm
pV = n RT
3 * 5 = n₁ * 0.0821 * 298
n₁ = ( 3 * 5 ) / ( 0.0821 * 298 ) = 0.6131
Pressure = 2.5 atm
pV = n RT
2.5 * 5 = n₁ * 0.0821 * 298
n₁ = ( 2.5 * 5 ) / ( 0.0821 * 298 ) = 0.5109
Diffrence in mole = 0.6131 - 0.5109 = 0.1022 mol
mole = mass in gram / molar mass
0.1022 = mass in gram / 16
Mass of methane gas leaked = 1.632 g
Ideal gas equation states,
pV = n RT
V = n RT / p
= 0.1022 * 0.0821 * 298 / 1
= 2.50 L
Therefore, the volume leaked = 2.5 L