Entropy increase when heating a gas
With constant volume heat capacity, it is defined as, for an ideal gas, that has the internal energy equation
$$dU=TdS-\rho dV= \frac{\nu}{2} Nk_{\beta} T\ (1)$$ which is the Ideal Gas Law, obtained from the first law of thermodynamics. Using moles, (1) is equivalent to
$$ \frac{\nu}{2}nRT $$
Now, onto heat capacities. 
$C_V$ is defined as,
$$C_V=(\frac{\partial U}{\partial T})_V= \frac{\partial}{\partial T}\left(\frac{\nu nRT}{2}\right)=\frac{\nu}{2} nR $$

Constant pressure heat capacity is defined as
$$C_{\rho}=C_V + nR=\frac{\nu+2}{2} nR$$, where $n$ is the amount of substance and $R$ is the universal gas constant. $\nu$ is the degrees of freedom of the molecule, which is easily obtained by thinking about a monoatomic ideal gas and the ways it can move. 

Below is some fun theory surrounding the heat capacity for constant volume.
Note that constant-pressure specific heat is also defined as
$$C_{\rho}=(\frac{\partial U}{\partial T})_{\rho}=dU+\rho dV+Vd\rho=
(\frac{\partial H}{\partial T})  $$
Where $H$ is the enthalpy of heating the gas, which is 
$$\mathrm{d}H=\mathrm{d}U+p\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S-\rho\mathrm{d}V+\rho\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S+V\mathrm{d}\rho$$
Enthalpy is commonly defined with a constant pressure process, thus $\mathrm{d}\rho=0$.
So, for a constant pressure, we have $$\mathrm{d}H=T\mathrm{d}S$$,
exploting the definition of entropy for a reversible change,
$$\mathrm{d}S=\frac{\mathrm{d}Q}{T} \rightarrow \mathrm{d}Q=T\mathrm{d}S $$
so, we finally get $$\mathrm{d}H=\mathrm{d}Q$$
which is a standard thermodynamic identity, and can also be used to get to the answer at the beginning of the post (though it is admittedly longer and more complicated to do so.)

Hope these hints help you!