Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 00 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 620 C. What is the new value of the pressure inside the bulbs ? The volume of the connecting tube is negligible.
When the bulbs are maintained at two different temperatures.
The total heat gained by 'B' is the heat lost by 'A'. Let the final temperature be X
So, m1SDt=m2SDt
⇒n1X=62n2−n2X
So, X = 310 C = 304K
[Since, Initial temperature = 00 C]
P = 76 cm of Hg, V1=V2Hencen1=n2]
for a single ball
P1V1T1=P2V2T2
⇒76×V273=P2×V304
⇒P2=304×76273
= 84.630 = 840 C