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Two cylinder having m(1)g and m(2)g of ...

Two cylinder having `m_(1)g` and `m_(2)g` of a gas at pressure `P_(1)` and `P_(2)` respectively are put in communication with each other, temperature remaining constant. The common pressure reached will be

A

`(P_(1)P_(2)(m_(1)+m_(2)))/(P_(2)m_(1)+P_(1)m_(2))`

B

`(P_(1)P_(2)m_(1))/(P_(2)m_(1)+P_(1)m_(2))`

C

`(m_(1)m_(2)(P_(1)+P_(2)))/(P_(2)m_(1)(P_(1)+P_(2)))`

D

`(m_(1)m_(2)P_(2))/(P_(2)m_(1)+P_(1)m_(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
According to Boyle's law, PV=kT (a constnt)
or `P(m)/(rho)kTor rho=(p m)/(kT)`
`or rho=(P)/(K)" ""("where (kT)/(m)=K a "constant"")"`
So, `rho1=(P_(1))/(K)and V_(1)(m_(1))/(rho1)=(m_(1))/(P_(1)//K)=(km_(1))/(P_(1))`
Similarly, `V_(2)=(Km_(2))/(P_(2))`
Total volume `=V_(1)+V_(2)=K((m_(1))/(P_(1))+(m_(2))/(P_(2)))`
Let P be the common pressure and `rho` be the common density of mixture. Then.
`rho=(m_(1)+m_(2))/(V_(1)+V_(2))=(m_(1)+m_(2))/(K((m_(1))/(P_(1))+(m_(2))/(P_(2))))`
`thereforeP=Krho=(m_(1)+m_(2))/((m_(1))/(P_(1))+(m_(2))/(P_(2)))=(P_(1)P_(2)(m_(1)+m_(2)))/((m_(1)P_(2)+m_(2)P_(1)))`
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