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Specific heat of argon at constant press...

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)`.

Text Solution

Verified by Experts

The correct Answer is:
`1.77`
Here `C_p = 0.125 "cal g"^(-1)K^(-1),`
`C_v = 0.075 "cal g"^(-1)K^(-1)`,
`J = 4.18 xx10^7 "erg cal"^(-1)`
Standard pressure, `P=1.01xx10^6 "dyne cm"^(-2)`
For 1g of a gas, `C_P - C_V = R/J`
`:. r = J (C_p - C_v)=4.18xx10^(7)xx(0.125 – 0.075)`
`= 4.18xx10^7 xx0.050 = 2.09 xx 10^(7) "erg g"^(-1)K^(-1)`
But for 1 g of a gas, PV =RT
`:.` Volume of 1 g of argon at S.T.P. is `V = (RT)/P = (2.09 xx 10^7 xx 273)/(1.01 xx10^6) =564.92 cm^3 `
Density of argon, `rho = ("Mass")/("Volume") =1/(564.92)= 1.77xx10^(-3)"gcm"^(-3)`
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