# Thermodynamics Heat flow into Ideal Gas

I'm currently looking at University Physics by Young and Freedman, and I came across a problem that is stumping me.

19.25. Heat $Q$ flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas.

The answer given by the book is $\frac 25$. This answer can be achieved when you note that the if $Q$ is completely turned to heat, we have $Q = nC_p\Delta T = \frac{5}{2}nR\Delta T$ and the amount of work done is $p\Delta V = nR\Delta T$, so the ratio is $\frac{2}{5}$.

However, what confuses me is that $Q$ is not completely turned to heat, but also is used to do work. When we try to divide the heat to its components, it seems to give the same answer. Suppose that $x$ is the answer. Then $xQ$ is converted to work and $(1-x)Q$ is converted to heat. Then the total energy is $$Q = xQ + (1-x)Q = nR\Delta T + \frac{3}{2} nR\Delta T = \frac{5}{2} nR\Delta T$$ This implies that $Q$, whether or not it is completely turned to work or heat, will always be $nC_p \Delta T$ as long as pressure is constant. Is this correct? This seems pretty weird.

I think you are confusing heat Q with change in internal energy U. The total heat supplied Q goes into increasing the internal energy by an amount of $\dfrac{3}{2} nR \Delta T$ and work done by the gas which is $nR \Delta T$ as you correctly said, which makes it a total of $\dfrac{5}{2} nR \Delta T$. This is always true for an ideal gas performing an isobaric(constant pressure) process. In fact, $C_p$ is defined to be the rate of heat flow per unit temperature per mole for a constant pressure process. Hence, $\dfrac{dQ}{dT}=n C_p$ in all such processes.