5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If $$C_V = 28$$ J K$$^{-1}$$, calculate $$\Delta U$$ and $$\Delta pV$$ for the process. (R = 8.0 J K$$^{-1}$$ mol$$^{-1}$$)

For any ideal gas, the change in its internal energy depends only on temperature. The quantitative relation is given by the formula

$$\Delta U = n\,C_V\,\Delta T,$$

where $$n$$ is the number of moles, $$C_V$$ is the molar heat capacity at constant volume, and $$\Delta T$$ is the change in temperature.

We have $$n = 5\ \text{mol},\; C_V = 28\ \text{J K}^{-1}\!\!\text{mol}^{-1},\; T_1 = 100\ \text{K},\; T_2 = 200\ \text{K}.$$ Hence

$$\Delta T = T_2 - T_1 = 200\ \text{K} - 100\ \text{K} = 100\ \text{K}.$$

Substituting the values,

$$\Delta U = 5 \times 28\ \text{J K}^{-1}\!\!\text{mol}^{-1} \times 100\ \text{K}.$$

Multiplying step by step, we get

$$28 \times 100 = 2800\ \text{J mol}^{-1},$$

and then

$$5 \times 2800\ \text{J mol}^{-1} = 14000\ \text{J}.$$

Converting joules to kilojoules gives

$$\Delta U = 14000\ \text{J} = 14\ \text{kJ}.$$

Next we evaluate the change in the quantity $$pV$$. For an ideal gas we have the equation of state

$$pV = nRT,$$

so any change in $$pV$$ is governed by

$$\Delta(pV) = nR\Delta T,$$

with the same $$\Delta T$$ as before. Taking $$R = 8.0\ \text{J K}^{-1}\!\!\text{mol}^{-1}$$, we substitute:

$$\Delta(pV) = 5 \times 8.0\ \text{J K}^{-1}\!\!\text{mol}^{-1} \times 100\ \text{K}.$$

First multiply the constants:

$$8.0 \times 100 = 800\ \text{J mol}^{-1},$$

then

$$5 \times 800\ \text{J mol}^{-1} = 4000\ \text{J}.$$

Changing units,

$$\Delta(pV) = 4000\ \text{J} = 4\ \text{kJ}.$$

We have obtained

$$\Delta U = 14\ \text{kJ}, \qquad \Delta(pV) = 4\ \text{kJ}.$$

Hence, the correct answer is Option C.