Question

Calculate $C_P$ for air, given that $C_v$ = 0.162 cal/g/K and density of air at N.T.P. is 0.001293 g/$cm^3$.

Answer 1

Hello frd ^_^

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$\bold\pink{Solution:}$

Given: $C_v$ = 0.162 cal/g/K

Dencity of air at N.T.P. = 0.001293 g/$cm^3$

To find : $C_P$= ?

Now, $C_P$-$C_v$=$\frac{r}{j}$

=>$C_P$-$C_v$=$\frac{PV}{Tj}$

(⚫.⚫PV = nRT)

=> $C_P$-$C_v$= $\frac{P.1}{d.TJ}$ (d=Dencity)

=>$C_P$-$C_v$= $\frac{1.01×10^6}{273×4.2×10^7×1.293×10^{-3}}$

=> $C_P$-$C_v$=$\frac{1.01×10^{2}}{1482.5}$

=> $C_P$-$C_v$ =$6.8×10^{-2}$

=> $C_P$-$C_v$= 0.068

=> $C_P$= 0.162 + 0.068

=> $C_P$= 0.23Cal$g^{-1}k{-1}$

Thank you ☺

By @Swigy

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Answer 2

Solution :

Given,

Constant volume = 0.162 cal/g/K

Density of air at N.T.P. = 0.001293 g/cm^3

For finding the constant pressure of air,

By using this formula,

Cp - Cv = r / J

=> Cp - Cv = PV / Tj

°.° PV = n × R × T

* Cp means Constant pressure.

* Cv means Constant volume.

=> Cp - Cv = P × 1 / p × Tj

(where, p = density)

=> Cp - Cv = 1.01 × 10^6 / 273 × 4.2 × 10^7 × 1.293 × 10^-3

=> Cp - Cv = 1.01 × 10^2 / 1482.5

=> Cp - Cv = 6.8 × 10^-2

=> Cp - Cv = 0.068

=> Cp = 0.162 + 0.068

=> Cp = 0.23 cal/g/K

Thus, the constant pressure of air is 0.23 cal/g/K.

Answer : Thus, the constant pressure of air is 0.23 cal/g/K.