The change in the internal energy of an ideal gas does not depend on?

One mole of an ideal diatomic gas undergoes a transition from \(A\) to \(B\) along a path \(AB\) as shown in the figure. 
         
The change in internal energy of the gas during the transition is:

The pressure in a diatomic gas increases from ${\mathrm{P}}_0$ to $3{\mathrm{P}}_0$, when its volume is increased from ${\mathrm{V}}_0$ $\mathrm{to}$ $2{\mathrm{V}}_0$. The increase in internal energy  will be:
     
1. $6{\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_0$

2. $8.5{\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_0$

3. $12.5{\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_0$

4. $14.5{\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_0$

A gas mixture consists of \(2\) moles of \(\mathrm{O_2}\) and \(4\) moles of \(\mathrm{Ar}\) at temperature \(T.\) Neglecting all the vibrational modes, the total internal energy of the system is:

The value \(\gamma = \frac{C_P}{C_V}\) for hydrogen, helium, and another ideal diatomic gas \(X\) (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to: