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Two flasks X and Y have capacity 1 litre and 2 litre respectively. Each of them contains 1 mole of a gas. The temperature of the flask are so adjusted that the average speed of molecules in X is twice as that in Y. The pressure in flask X would be (A) 8 times of that in Y (B) twice of that in Y (C) same as that in Y (D) half of that in Y

Two flasks X and Y have capacity 1 litre and 2 litre respectively. Each of them contains 1 mole of a gas. The temperature of the flask are so adjusted that the average speed of molecules in X is twice as that in Y. The pressure in flask X would be 
(A) 8 times of that in Y (B) twice of that in Y
(C) same as that in Y (D) half of that in Y

Grade:11

1 Answers

Susmita
425 Points
5 years ago
Let us denote the state of gas X by P1,V1,T1 and that of gas Y by P2,V2,T2.
Given V1=1 lt,n1=1 mole and V2=2lt,n2=1mole.
The average speed 
v=\sqrt{\frac{8kT}{\pi m}}
So
\frac{v1}{v2} =\sqrt{\frac{T1}{T2}}
Or,T1/T2=v12/v22
Given v1=2v2
So T1/T2=4
From ideal gas equation
P1V1=n1RT1 and P2V2=n2RT2
Dividing
P1V1/P2V2=n1RT1/n2RT2
Or,P1/P2=(V2/V1)(T1/T2)=2×4=8
So answer is A.please approve if helped.

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