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Question:

Two flasks X and Y have capacity 1L and 2L respectively and each of them contains 1 mole of a gas. The temperatures of the flasks are so adjusted that the average speed of molecules in X is twice as those in Y. The pressure in flask X would be:

same as that in Y

half of that in Y

twice of that in Y

8 times of that in Y

Solution:

According to Kinetic Theory of gases, PV = (1/3)mn<u2> where, P is pressure of the gas, m is mass of the gas, n is no. of molecules and u is the average speed of the molecules of the gas.
So, for gas present in flask X and Y will have the pressure as follows
Let PX and PY be the pressure in flask X and Y respectively.
Let nX and nY be the number of molecules in flask X and Y respectively.
Let uX and uY be the average speed of molecules in flask X and Y respectively.
Given that nX = nY = 1 mole
Given that uX = 2uY
Volume of flask X, VX = 1L
Volume of flask Y, VY = 2L
Then, PXVX = (1/3)m nX uX2
PYVY = (1/3)m nY uY2
Dividing the two equations:
(PXVX)/(PYVY) = (nXuX2)/(nYuY2)
Since nX = nY, we have:
(PXVX)/(PYVY) = (uX2)/(uY2)
(PX * 1)/(PY * 2) = (2uY)2/(uY2)
PX/(2PY) = 4
PX = 8PY
Therefore, the pressure in flask X would be 8 times that in Y.