2.2 g of nitrous oxide (N$$_2$$O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, $$\Delta$$U is '-x'J. The value of 'x' is ______.
[nearest integer]
(Given: atomic mass of N = 14 g mol$$^{-1}$$ and of O = 16 g mol$$^{-1}$$. Molar heat capacity of N$$_2$$O is 100 JK$$^{-1}$$ mol$$^{-1}$$)

We have 2.2 g of N$$_2$$O, at a constant pressure of 1 atm, cooled from 310 K to 270 K, and its volume changes from 217.1 mL to 167.75 mL.

The molar mass of N$$_2$$O is 2(14) + 16 = 44 g/mol, so $$n = \frac{2.2}{44} = 0.05 \text{ mol}$$.

Under constant pressure, the enthalpy change ($$\Delta H = q_p$$) is $$\Delta H = nC_p\Delta T = 0.05 \times 100 \times (270 - 310) = 0.05 \times 100 \times (-40) = -200 \text{ J}$$.

The work done by the system is $$P\Delta V = 1 \text{ atm} \times (167.75 - 217.1) \text{ mL} = 1 \times (-49.35) \text{ mL} \cdot \text{atm}$$, and converting gives $$P\Delta V = -49.35 \times 10^{-3} \times 101.325 \text{ J} \approx -5 \text{ J}$$.

Applying the first law of thermodynamics, $$\Delta U = \Delta H - P\Delta V = -200 - (-5) = -200 + 5 = -195 \text{ J}$$. Since $$\Delta U = -x$$ J, the value of $$x$$ is 195.