When to use C(subv) and C(subp) for Q heat equation

1. Nov 27, 2016

nso09

1. The problem statement, all variables and given/known data
A player bounces a basketball on the floor, compressing it to 80.5% of its original volume. The air (assume it is essentially N2 gas) inside the ball is originally at a temperature of 20.5°C and a pressure of 1.80 atm. The ball's diameter is 23.9 cm.
By how much does the internal energy of the air change between the ball's original state and its maximum compression?
Given: initial and final volume, initial temperature 20.5 degrees celsius, initial pressure 1.80 atm, diameter

2. Relevant equations
$Q=nC_v\Delta(T)$
$C_p=C_v+R$

3. The attempt at a solution
Q=0 since this seems to be an adiabatic process. Therefore $W=-\Delta(U)$
So $\Delta(U)=nC_v\Delta(T)$

My question is, why can't we use $C_p$ since the pressure is constant? I am confused on the cases where you have to use $C_p$ or $C_v$. Is this just because it is adiabatic? The volume seems to change though since they tell us that the ball compresses to 80.5% of its original volume. So why $C_v?$

2. Nov 27, 2016

TSny

$U$ is a state variable. So the change, $\Delta U$, when going from some initial state to some final state is independent of the type of process connecting the two states. For an ideal gas $U = nC_V T$, which expresses that $U$ is proportional to $T$. The proportionality constant happens to be $nC_V$. But the $C_V$ here is just a number and it does not imply that you must have a constant volume process in order to use $\Delta U = nC_V\Delta T$. This equation is valid for all processes for any ideal gas.

However, the heat $Q$ transferred in going from the initial state to the final state does depend on the particular process. $Q = nC_V \Delta T$ is used for a constant volume process, while $Q = nC_P \Delta T$ is for a constant pressure process.

3. Nov 27, 2016

Staff: Mentor

See the following: http://physics.stackexchange.com/qu...delta-e-int-nc-v-delta-t-for-an/295200#295200