Question 2: Thermodynamics & KTG

Question

If $$7 \mathrm{~gm} \mathrm{~N}_2$$ is mixed with $$20 \mathrm{~gm} \mathrm{~Ar}$$, there $$\frac{C_p}{C_V}$$ of mixture will be

Accepted Answer

For $$\mathrm{N}_2$$, degree of freedom $$f=5$$ [For di-atomic gas]
$$	herefore \quad C_{V_1}=\frac{f}{2} \cdot R=\frac{5}{2} R$$ and $$C_{p_1}=C_{V_1}+R=\frac{7}{2} R$$
Amount of $$\mathrm{N}_2$$ gas, $$m_1=7 \mathrm{~g}$$
$$	herefore$$ Number of moles of $$\mathrm{N}_2$$,
$$\begin{aligned}
& n_1=\frac{m_1}{28}=\frac{7}{28} \
& n_1=\frac{1}{4}
\end{aligned}$$
For Ar, degree of freedom, $$f=3$$
$$\begin{aligned}
& 	herefore \quad C_{V_2}=\frac{3}{2} R \
& 	herefore \quad C_{p_2}=C_{V_2}+R \
& =\frac{3}{2} R+R=\frac{5}{2} R \
&
\end{aligned}$$
Amount of Ar gas, $$m_2=20 \mathrm{~g}$$
$$	herefore$$ Number of moles of Ar gas,
$$\begin{aligned}
n_2 & =\frac{m_2}{40}=\frac{20}{40}=\frac{1}{2} \
	herefore \quad\left(C_pight)_{	ext {mix }} & =\frac{n_1 C_{p_1}+n_2 C p_2}{n_1+n_2} \
& =\frac{\frac{1}{4} 	imes \frac{7}{2} R+\frac{1}{2} 	imes \frac{5}{2} R}{\frac{1}{4}+\frac{1}{2}}=\frac{17 R}{6} \
\left(C_Vight)_{	ext {mix }} & =\frac{n_1 C_{V_1}+n_2 C_{V_2}}{n_1+n_2} \
& =\frac{\frac{1}{4} 	imes \frac{5}{2}+\frac{1}{2} 	imes \frac{3}{2} R}{\frac{1}{4}+\frac{1}{2}}=\frac{11 R}{6} \
	herefore \quad\left(\frac{C_p}{C_V}ight)_{	ext {mix }} & =\gamma_{	ext {mix }}=\frac{\frac{17 R}{6}}{\frac{11R}{6}}=\frac{17}{11}
\end{aligned}$$

Answer

$$\frac{17}{6}$$

Answer

$$\frac{11}{7}$$

Answer

$$\frac{17}{11}$$

Answer

$$\frac{17}{13}$$

Thermodynamics & KTG

If $7 \mathrm{~gm} \mathrm{~N}_2 is mixed with 20 \mathrm{~gm} \mathrm{~Ar}, there \frac{C_p}{C_V}$ of mixture will be

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