Question 90: Thermodynamics & KTG

Question

A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio $$C_p/C_V$$ of the mixture is

Accepted Answer

Mass of helium, $$m_1=16 \mathrm{~g}$$
Mass of oxygen, $$m_2=16 \mathrm{~g}$$
$$	herefore$$ Number of moles of He gas,
$$n_1=\frac{m_1}{4}=\frac{16}{4}=4$$
Number of moles of $$\mathrm{O}_2$$ gas,
$$n_2=\frac{m_2}{32}=\frac{16}{32}=\frac{1}{2}
$$
We know that for the mixture,
$$C_V=\frac{n_1 C_{V_1}+n_2 C_{V_2}}{n_1+n_2}, 	ext { where } C_V=\frac{f R}{2}
$$
and $$C_p=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}$$, where $$C_p=\left(\frac{f}{2}+1ight) R$$
For He gas, $$f=3, n_1=4$$
For $$\mathrm{O}_2$$ gas, $$f=5, n_2=\frac{1}{2}$$
$$\begin{aligned}	herefore \quad \frac{C_p}{C_V} & =\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1 C_{V_1}+n_2 C_{V_2}} \
& =\frac{4 	imes\left(\frac{3}{2}+1ight) R+\frac{1}{2}\left(\frac{5}{2}+1ight) R}{4 	imes \frac{3}{2} R+\frac{1}{2} 	imes \frac{5}{2} R} \
& =\frac{\left(4 	imes \frac{5}{2} Right)+\left(\frac{1}{2} 	imes \frac{7}{2} Right)}{\left(4 	imes \frac{3}{2} Right)+\left(\frac{1}{2} 	imes \frac{5}{2} Right)}=\frac{47}{29} \
& =1.62
\end{aligned}$$

Answer

1.4

Answer

1.54

Answer

1.59

Answer

1.62

Thermodynamics & KTG

A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio $C_p/C_V$ of the mixture is

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