Question

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)`.

Answer 1

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)`.