Answer : $\;C_{p}+C_{v}\;$ is larger for a diatomic gas than a monoatomic gas.

Explanation :

$C_{v} = \large\frac{f_{2}}{2}R$

$C_{p} = \large\frac{f+2}{f} R$

f is the no . of degrees of freedom.

$\quad\; \qquad$Monoatomic $\qquad \; \qquad \;$ $\qquad \qquad$ Diatomic

f $\quad\; \qquad$ 3 $\qquad \; \qquad \;$ $\qquad \qquad \qquad \qquad$ 5

$C_{v}$ $\quad\; \qquad$ =$\large\frac{3}{2}R$ $\qquad \; \qquad \;$ $\qquad \qquad \qquad \qquad$ $\large\frac{5}{2}R$

$C_{p}$ $\quad\; \qquad$ =$\large\frac{5}{2}R$ $\qquad \; \qquad \;$ $\qquad \qquad \qquad \qquad$ $\large\frac{7}{2}R$

$C_{p} -C_{v}$ $\quad\; \quad$ = R $\qquad \; \qquad \;$ $\qquad \qquad \qquad \qquad$ R

$C_{p} +C_{v}$ $\quad\; \quad$ = 4R $\qquad \; \qquad \;$ $\qquad \qquad \qquad \quad$ 6R

$C_{p}\;.C_{v}$ $\quad\; \qquad$ =$\large\frac{15}{4}R$ $\qquad \; \qquad \;$ $\qquad \quad \qquad \quad$ $\large\frac{25}{4}R$

$\large\frac{C_{p}}{C_{v}}$ $\quad\; \qquad$ =$\large\frac{5}{3}=1.6$ $\qquad \; \qquad \;$ $\qquad \quad \qquad \quad$ $\large\frac{7}{5}=1.4$