pressure in a mixture of 4 gram of O2 and 2 gram of H2 confined in a bowl of 1 litre at zero degree celsius is
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Answered by
30
Given:
V = 1L
T = 273K
Let us find the Partial pressure of O2.
No. of moles = 4/32 =0.125 moles
P1= nRT/v =0 .125x0.082 x 273/1 =2.80 atm
Now let us find the partial pressure of H2.
P2 = nRT/v =1 x 0.082 x 273/1 =22.386 atm
Total pressure of the mixture = 2.80 + 22.38 =25.18 atm
V = 1L
T = 273K
Let us find the Partial pressure of O2.
No. of moles = 4/32 =0.125 moles
P1= nRT/v =0 .125x0.082 x 273/1 =2.80 atm
Now let us find the partial pressure of H2.
P2 = nRT/v =1 x 0.082 x 273/1 =22.386 atm
Total pressure of the mixture = 2.80 + 22.38 =25.18 atm
Answered by
13
Total pressure is 22.18
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