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Calculate the molar specific heat at con...

Calculate the molar specific heat at constant volume . Given : specific heat of hydrogen at constant pressure is `6.85 "cal" mol^(-1) K^(-1)` and density of hydrogen = `0.0899 "g" cm^(-3)`. One mole of gas =`2.016g, J=4.2xx10^(7) "erg" cal^(-1)` and `1` atmosphere = `10^(6) "dyne" cm^(-2)`.

Text Solution

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Since the density of hydrogen is `0.0899 g cm^(-3)` therefore volume occupied by `0.0899 `g of hydrogen at `NTP` is `1cm^(3)` . So, volume of `1` mole `(2.016g)` of gas ,`V=(1)/(0.0899)xx2.016 cm^(3)`
`C_(P) - C_(V) = (R)/(J) = (PV)/(TxxJ) = (10^(6) xx 1xx 2.016)/(0.0899xx273xx4.2xx10^(7))= 1.96xx10^(-3) "cal" mol^(-1)K^(-1)`
` therefore C_(V) -C_(P) - 1.96 = (6.85 -1.96)= 4.89 "cal" mol^(-1) K^(-1)`
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