Chemistry, asked by einstien5863, 11 months ago

31 A 2L vessel contains 4 g of Helium and 4 g of He
gas at 27°C. After sometime, 50% of the gas having
higher average speed is removed. What is the
percentage reduction in total pressure if temperature
remains constant?
(1) 66.67%
( 33.33%
(3) 50%
(4) 25%​

Answers

Answered by antiochus
24

Answer:

Volume of vessel=2L

He=4g

n_{He} =\frac{4}{4} =1mol

H2=4g

n_{H} =\frac{4}{2} =2mol

Total moles=2+1=3mol

T=27C=300K

P=\frac{nRT}{V}

P=\frac{3*0.0820*300}{2L}

P=36.90atm

50% of the gas having higher average speed is removed.

P=\frac{(1*1)*0.082*300}{2}

P=24.6atm

Reduction in pressure=36.9-24.6

                                    =12.3

Then the reduction % will be \frac{12.3}{36.9} *100

=33.3%

option 2 is correct

Answered by callatripatel
0

""CORRECT EXPLANATION """"

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