Question

5 moles of an ideal monoatomic gas absorbs '$X$' joule heat when heated from ${25}^{\circ }C$ to ${30}^{\circ }C$ at constant volume. The amount of heat absorbed when 2 moles of the same gas is heated from ${25}^{\circ }C$ to ${30}^{\circ }C$ at constant pressure , is:

• A. $\frac{3}{5}XJ$
• B. $\frac{5}{3}XJ$
• C. $\frac{2}{3}XJ$
• D. $\frac{25}{6}XJ$

Answer 1

The correct option is C $\frac{2}{3}XJ$

For ideal monatomic gas : $C_v=\frac{3}{2}R\text{We know}{C}_p-C_v=R\therefore C_p=C_v+R=\frac{3}{2}R+R=\frac{5}{2}R$
$\because$ We know that $q_v=\Delta u=n{C}_{v}dT$
$q_v=n{C}_{v}dT$

Again, according to the question, $q_v=X$
$\therefore X=5\times \frac{3}{2}R$
Heat capacity at constant pressure = $\frac{15}{2}R+5R=\frac{25}{2}R$
Heat capacity at constant pressure for 2 moles $\frac{25\times 2}{2\times 5}R=5R$

ratio of Heat capacity at constant pressure of 2 moles to the heat capacity at constant volume of moles $=\frac{5R}{X}=\frac{5R\times 2}{15R}=\frac{2}{3}$
Heat capacity at constant pressure of 2 moles = $\frac{2}{3}XJ$