Question

A gas at $50°\mathrm{C}$ constant pressure and a constant pressure requires the heat $160\mathrm{calories}$ to raise the temperature. The same mass of gas requires $100°\mathrm{C}$ a constant volume and $240\mathrm{calories}$ heat is released. Degrees of freedom does each molecule of this gas have? (Assume that gas is ideal)

• A. $7$
• B. $5$
• C. $6$
• D. $3$

Answer 1

The correct option is C

$6$

Explanation for the correct options

(C) $6$

Step 1:

• ${\mathrm{Q}}_{\mathrm{p}}=160\mathrm{cal}$
• where ${\mathrm{Q}}_{\mathrm{p}}$ is heat transfer at constant pressure
• ${\mathrm{nC}}_{\mathrm{p}}∆\mathrm{T}=160$
• where ${\mathrm{C}}_{\mathrm{p}}$ is molar-specific heat, $\mathrm{T}$ is temperature, and $\mathrm{n}$ is moles
• Applying the above formula
• $160={\mathrm{nC}}_{\mathrm{p}}50—\left(\mathrm{i}\right)$

Step 2:

• ${\mathrm{Q}}_{\mathrm{v}}=240\mathrm{cal}$
• where ${\mathrm{Q}}_{\mathrm{v}}$ is the heat transfer at a constant volume
• $240={\mathrm{nC}}_{\mathrm{v}}100—\left(\mathrm{ii}\right)$

Step 3:

• From equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$
• $\frac{160}{240}=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}\times \frac{1}{2}$
• $\begin{array}{rcl}\mathrm{\gamma }& =& \frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}\\ & =& 1+\frac{2}{\mathrm{f}}=\frac{4}{3}\\ \mathrm{f}& =& \frac{2}{\mathrm{\gamma }-1}\\ & =& \frac{2}{{\displaystyle \frac{4}{3}-1}}\\ \mathrm{f}& =& 6\end{array}$

Final Answer: Hence option (C) is correct.