Entropy increase when heating a gas What is the entropy change when the temperature of 1.00 mol of ideal gas is raised reversibly from 10 °C to 50 °C at constant pressure, constant volume, and assuming the heat capacity is independent of temperature?

This is all the information given. I am assuming that the formula $\Delta S = C \ln (T_2/T_1)$ is used. However, I do not know how the heat capacities are supposed to be obtained from the given information. I do not have the enthalpy or internal energy of the process. I cannot calculate the heat capacity from pressure or volume. 
Please say if my reasoning is incorrect.
(These aren’t the values of the original question but this is based on Atkins Chemical Principles 5th ed. Ch 8 # 5 and 6)
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!