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Question

A container (volume V) has an ideal gas at 4 atm and 300 K. This is connected by a tube with another container of volume 2 V, which has a gas at 8 atm and 600 K. What will be final pressure of each container if the temperature is maintained - the container with volume V at 300 K and the 2V one at 600 K?

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Solution

When the tubes are connected, pressures become the same.
Also, the total number of moles before and after the connection are the same.
n1=4V300R,n2=16V600R,

total moles = n1+n2=V25R


n1=n2=2n=V25R

n=V50R

When the tube is connected:
P1=P2 =P

PV=nRT

P=nRTV ; (n=V50R )

So, P = V50R×300RV=6 atm


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