Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0°C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62°C. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.
Given
Initial temperature of gas in both bulbs T1 =0
Initial pressure of gas in both bulbs P1=P2=76cm of Hg=0.76m of Hg
Temperature of bulb placed in ice T2=0
=273.15K
Temperature of other bulb T’2=62
=335.15K
Let each of the bulbs have n1 moles initially.
Now since the second bulb has kept at higher temperature gas in second bulb will expand and some of the gas will flow to first bulb and number of moles will change in second bulb.
Let the number of moles in second bulb after its pressure reached P be n2.
Volume of both the bulbs is same so V1=V2=V.
Applying ideal gas equation in both bulbs
PV=nRT
Where V= volume of gas
R=gas constant =8.3JK-1mol-1
T=temperature
n=number of moles of gas
P=pressure of gas.
and equating the value of constant R
Number of moles left out of second bulb after temperature rose=n1-n2
Let n3 moles be left when pressure reached P in fist bulb. Applying ideal gas equation in first bulb, before and after temperature change.
Also,
=own moles of first bulb n1+moles received from second bulb
=n1+n1-n2
Substituting the value of n3 and n2 in above equation
Taking n1 common
P=0.8375m of Hg
The new value of the pressure inside the bulbs is 83.75cm of Hg.