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 76cm 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 neigligible.

Explanation

When the bulbs are maintained at two
different temperature .
The total heat gaine by 'b' is the heat lost
by 'A'. Let the final temperature be X
So, ` m_1 SDt= m_2 SDt `
` rArr (n_1)M xx S xx (X-0)= (n_2)M xx S xx (62 - X)`
` rArr (n_1)X = 62(n_2)- (n_2)X`
` So, X = 31^@C = 304K`
` [Since , Initial temperature = 0^@C `
` P = 76 cm of Hg, V_1 =V_2 Hence n_1 = n_2]`
for a single ball
` ((P_1)(V_1)/(T_1))=((P_2)(V_2)/(T_2))`
` rArr (76 xx V/ 273) = ((P_2) xx V/304) `
` rArr (P-2) = (304 xx 76 /273)`
` = 84.630 = 84^@C` .