Question

$\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}=\frac{5}{3}$two moles of an ideal gas are mixed with three moles of another ideal gas$\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}=\frac{4}{3}$. $\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}$for the mixture is?

• A. $1.47$
• B. $1.4$
• C. $1.42$
• D. $1.50$

Answer 1

The correct option is C

$1.42$

Explanation for the correct options:

C. $1.42$

Step1:

• $\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}$ is known as the heat capacity ratio.
• It is defined as the ratio of the constant pressure of heat capacity to the constant volume.
• $\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}={\mathrm{\gamma }}_{\mathrm{eq}}$--(i)

Step2:

• ${\mathrm{C}}_{{\mathrm{p}}_{\mathrm{eq}}}=\frac{{\mathrm{n}}_1{\mathrm{CP}}_1+{\mathrm{n}}_2{\mathrm{CP}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}$
• ${\mathrm{C}}_{{\mathrm{v}}_{\mathrm{eq}}}=\frac{{\mathrm{n}}_1{\mathrm{Cv}}_1+{\mathrm{n}}_2{\mathrm{Cv}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}$

Step3:

• ${\gamma }_{eq}=\frac{2\times {\displaystyle \frac{5R}{2}}+3\times {\displaystyle \frac{8R}{2}}}{2\times {\displaystyle \frac{3R}{2}}+3\times {\displaystyle \frac{6R}{2}}}$
• $=\frac{5+12}{3+9}$$=\frac{17}{12}=1.42$

Therefore,$\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}$ or ${\gamma }_{eq}$ for the mixture$\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{42}$