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Two glass bulbs of equal volume are conn...

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at `0^(@)C` and 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.

Text Solution

Verified by Experts

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` .
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