Entropy change Carnot-process

Say that an ideal monatomic gas undergoes a reversible Carnot-process between to temperatures with the difference $\Delta T$ such that the quotient between temperatures is $q$. Since only the isothermal processes change the entropy we should be able to calculate it by integrating the following: $$dS = \frac{1}{T}(vC_V dT + pdV) = \frac{1}{T} dV\frac{vRT}{V} = vR \frac{dV}{V}$$ We get: $$S = \int_{\Delta T} vR \frac{dV}{V} = vRln(q)$$ Have I understood the concept of entropy change in a Carnot-process correctly? The reason I ask is because I got an incorrect answer on a question that was specified as:

1 kilomole of an ideal monatomic gas undergoes a reversible Carnot-process between the temperatures $T_1 = 573 K$ and $T_2 = 293 K$. The work done by the gas during the process is 1500 kJ. Calculate the change of entropy during the partial processes.

And so with my formula I get $S = 1000 \cdot 8.3145 \cdot ln(573/293) \approx 5.58 kJ$ but the answer sheet says $5.36 kJ$. So have I misunderstood anything or what's wrong?