InterviewSolution
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InterviewSolution
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Specific heat of argon at constant pressure is `0.125 cal.g^(-1) K^(-1)`, and at constant volume `0.075 cal.g^(-1)K^(-1)`. Calculate the density of argon at N.T.P. Given `J = 4.18 xx 10^(7)" erg "cal^(-1)` and normal pressure = `1.01 xx 10^(6)" dyne "cm^(-2)`. |
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Answer» Here, `C_(p) = 0.125 cal. G^(-1) K^(-1)`, `C_(upsilon) = 0.075 cal. G^(-1) K^(-1)` , `J = 4.18 xx 10^(7)erg cal.^(-1)` Normal pressure , `P = 1.01 xx 10^(6)dyn e//cm^(2)` Normal temperature , `T = 273^(@)K` we have to calculate density , `rho =?` As, gas constant for 1 g of gas `r = (P upsilon)/(T) = (P)/(rho T)` Where, `upsilon` = volume oc cupied by 1 gram of gas Now, `C_(p) - C_(upsilon) - r/J` or, `r = (C_(p)-C_(upsilon))J` `(P)/(rho T) = (0.125 - 0.075) 4.18 xx 10^(7)` or `(1.01 xx 10^(6))/(rho xx 273) = 0.05 xx 4.18 xx 10^(7)` `:. rho = (1.01 xx 10^(6))/(273 xx 0.05 xx 4.18 xx 10^(7))` `=1.77 xx 10^(-3) g cm^(-3)`. |
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