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 negligible.

Answer

83,83

Explanation

`1/2 ("m"_("gas")) v_0^2 = nC_V Delta T" ":." "V_0 = sqrt((2nC_V Delta T)/("m"_("gas"))) = sqrt(((2)(n)(5/2xx8.31)(1))/((n)(32xx10^(-3))))="3 m/s"`Let x moles transfer from bulb of higher temperature to lower temperature.
Applying for both the bulbs initially and finally
`76 xxV = nR xx273 " ...(i)"`
` pxxV = (n+x)R 273" ....(ii)"`
`p xx V = (n-x) R xx 335 " ...(iii)"`
Solving these equations we get `n = 602/62 xx`
Now, dividing Eq. (ii) with Eq (i), we get
`(P')/76 =(n+x)/n =1+x/n = 1+ 62//602 = 1.103 " ":." "p' = 83.83"cm of Hg"`