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Question
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.
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Solution
Here ,
`P_1 = 0.76 \text { m Hg }`
`P_2 = P`
`T_1 = 273 K`
`T_2 = 335 K`
Let each of the bulbs have `"n"_1` moles initially .
Let the number of moles left in second bulb after its pressure reached P be `"n"_2`.
Applying equation of state , we get
`(P_1V)/(n_1T_1)` = `(PV)/(n_2T_2)`
⇒ `0.76/(273n_1)` = `P/(335n_2)`
⇒ `n_2` = `(273P)/(335 xx 0.76)n_1`
Number of moles left in the second bulb after the temperature rose = `n_1 - n_2`
= `n_1 - (273P)/(335 xx 0.76) n_1`
Let `"n"_3` moles be left when pressure reached P . Applying equation of state in the first bulb , we get
`(P_1V)/(n_1T_1) = (PV)/(n_3T_1)`
⇒ `0.76/n_1 = P/n_3`
⇒ `n_3 = (Pn_1)/0.76`
`"n"_3` = its own `n_1` moles + the it received from the first
`"n"_3 = n_1+(n_1 - n_2)`
⇒ `(Pn_1)/0.76` = `"n"_1 + n_1` - `(273P)/(335 xx 0.76 )n_1`
⇒ `P/0.76 = 2 - (273P)/(335 xx 0.76)`
⇒ P = 0.8375
⇒ P = 84 cm of Hg
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