When $2moles$ of an ideal monoatomic gas are heated from $300K$ to $600K$ at constant pressure. The change in entropy of gas $\left(△S\right)$ is:

The correct option is C $5R\ln 2$
The entropy change for an isobaric process is given as:
$△S=n{C}_{p,m}ln\frac{T_2}{T_1}$
where
$T_2=600K$
$T_1=300K$
$n=2$
for ideal monoatomic gas we know
$C_{p,m}=\frac{5}{2}R$

Putting the values we get:
$△S=2\times \frac{5}{2}R\times \ln \frac{600}{300}$
= $5R\ln 2$