An engine takes in 5 moles of air at 20 $$°$$C and 1 atm, and compresses it adiabatically to $$1/10^{th}$$ of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is:

We are given that 5 moles of air (treated as an ideal di-atomic gas with rigid molecules) are initially at a temperature of $$T_i = 20^{\circ}\text{C} = 293\ \text{K}$$ and a pressure of $$P_i = 1\ \text{atm} = 1.013\times10^{5}\ \text{Pa}$$. The gas is compressed adiabatically to one-tenth of its original volume, i.e. $$V_f = \dfrac{V_i}{10}$$. We have to find the change in internal energy $$\Delta U$$ produced by this adiabatic compression.

For an ideal gas, the internal energy depends only on temperature, and the change in internal energy is given by the formula

$$\Delta U = n\,C_v\,(T_f - T_i)$$

where $$n$$ is the number of moles, $$C_v$$ is the molar heat capacity at constant volume, $$T_i$$ is the initial absolute temperature and $$T_f$$ is the final absolute temperature. Hence our first task is to determine $$C_v$$ and $$T_f$$.

Because the molecules are rigid di-atomic, they possess three translational and two rotational degrees of freedom, so the total degrees of freedom are $$f = 5$$. The molar heat capacity at constant volume is therefore

$$C_v = \dfrac{f}{2}\,R = \dfrac{5}{2}\,R$$

where $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$$ is the universal gas constant. Thus

$$C_v = \dfrac{5}{2}\,(8.314) = 20.785\ \text{J mol}^{-1}\text{K}^{-1}.$$

For an adiabatic (reversible) process in an ideal gas we have the relation

$$T\,V^{\gamma - 1} = \text{constant},$$

where $$\gamma = \dfrac{C_p}{C_v} = 1 + \dfrac{2}{f}.$$ Substituting $$f = 5$$ gives

$$\gamma = 1 + \dfrac{2}{5} = \dfrac{7}{5} = 1.4.$$

Using the adiabatic relation for the initial state $$(T_i,V_i)$$ and the final state $$(T_f,V_f)$$ we write

$$T_i\,V_i^{\gamma-1} = T_f\,V_f^{\gamma-1}.$$

Dividing both sides by $$T_i$$ and rearranging for $$T_f$$ we get

$$T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma-1}.$$

Since $$V_f = \dfrac{V_i}{10}$$, the volume ratio is

$$\dfrac{V_i}{V_f} = 10.$$

Therefore

$$T_f = 293\ \text{K}\; \times\; 10^{\,\gamma-1} = 293\ \text{K}\; \times\;10^{\,1.4 -1} = 293\ \text{K}\; \times\;10^{0.4}.$$

Now, $$10^{0.4} = e^{\,0.4\ln 10} = e^{\,0.4 \times 2.302585} = e^{0.921034} \approx 2.512.$$ Hence

$$T_f \approx 293\ \text{K}\,\times\,2.512 \approx 737\ \text{K}.$$

The change in temperature is therefore

$$\Delta T = T_f - T_i = 737\ \text{K} - 293\ \text{K} = 444\ \text{K}.$$

Substituting $$n = 5$$, $$C_v = 20.785\ \text{J mol}^{-1}\text{K}^{-1}$$ and $$\Delta T = 444\ \text{K}$$ into the formula for $$\Delta U$$, we obtain

$$\Delta U = 5\ \text{mol}\;\times\;20.785\ \dfrac{\text{J}}{\text{mol K}}\;\times\;444\ \text{K}.$$ $$\Delta U = 5 \times (20.785 \times 444)\ \text{J}.$$

First compute the bracketed product:

$$20.785 \times 444 = 20.785 \times (400 + 44) = 20.785 \times 400 + 20.785 \times 44 = 8\,314 + 914.5 = 9\,228.5\ \text{J}.$$

Now multiply by 5 moles:

$$\Delta U = 5 \times 9\,228.5\ \text{J} = 46\,142.5\ \text{J} \approx 4.614\times10^{4}\ \text{J}.$$

Converting to kilojoules,

$$\Delta U \approx 46.1\ \text{kJ}.$$

Rounded to the nearest integer, $$\Delta U = 46\ \text{kJ}.$$ So, the answer is $$46\ \text{kJ}$$.