Latent heat of vaporization of a liquid at 500 K and at 1 atm pressure is 10 kcal/mol. What will be the change in internal energy (ΔU) of 3 moles of liquid at the same temperature?
(A) 13 kcal
(B) -13 kcal
(C) 27 kcal
(D) -27 kcal

To solve this problem, we need to use the relationship between enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta U\)) for phase transitions like vaporization.

### Given:
- Latent heat of vaporization (\(\Delta H_{\text{vap}}\)) = 10 kcal/mol
- Temperature (\(T\)) = 500 K
- Number of moles (\(n\)) = 3
- Pressure (\(P\)) = 1 atm

We know that the relationship between the change in enthalpy (\(\Delta H\)) and the change in internal energy (\(\Delta U\)) is given by:

\[
\Delta H = \Delta U + P \Delta V
\]

Where:
- \(P\) is the pressure.
- \(\Delta V\) is the change in volume during vaporization.

Since the volume of the gas is much larger than that of the liquid, the change in volume is dominated by the volume of the gas. For an ideal gas, the volume can be expressed using the ideal gas law:

\[
PV = nRT
\]

So, the change in volume during vaporization (\(\Delta V\)) can be approximated as:

\[
\Delta V = \frac{nRT}{P}
\]

Now, we can rewrite the equation for \(\Delta U\):

\[
\Delta U = \Delta H - P \Delta V
\]

Substitute \(\Delta V = \frac{nRT}{P}\):

\[
\Delta U = \Delta H - P \left(\frac{nRT}{P}\right)
\]

Simplifying this:

\[
\Delta U = \Delta H - nRT
\]

Now, let’s substitute the values:
- \(\Delta H = 10\) kcal/mol (latent heat of vaporization per mole),
- \(n = 3\) moles,
- \(R = 2\) cal/mol·K (gas constant in kcal/mol·K),
- \(T = 500\) K.

First, calculate \(nRT\):

\[
nRT = 3 \times 2 \times 500 = 3000 \, \text{cal} = 3 \, \text{kcal}
\]

Now, use this to find \(\Delta U\):

\[
\Delta U = 10 \times 3 - 3 = 30 - 3 = 27 \, \text{kcal}
\]

### Conclusion:
The change in internal energy is 27 kcal. Hence, the correct answer is **(C) 27 kcal**.