1. Given data

• Mass of $N_2O = 2.2\,\text{g}$
• Molar mass of $N_2O = 2(14)+16 = 44\,\text{g mol}^{-1}$
• Hence moles, $n = \frac{2.2}{44} = 0.05\,\text{mol}$
• Temperature changes from $310\,\text{K}$ to $270\,\text{K}$ $\Delta T = 270-310 = -40\,\text{K}$
• Pressure is constant at $1\,\text{atm}$
• Molar heat capacity given: $C_p = 100\,\text{J K}^{-1}\text{mol}^{-1}$

2. Heat absorbed/released at constant pressure

At constant pressure, $q_p = nC_p\Delta T$ So, $q_p = 0.05 \times 100 \times (-40) = -200\,\text{J}$

Thus, the gas releases $200\,\text{J}$ of heat.

3. Work done

At constant external pressure, $w = -P\Delta V$ Here,

• Initial volume $V_i = 217.1\,\text{mL}$
• Final volume $V_f = 167.75\,\text{mL}$

So, $\Delta V = V_f - V_i = 167.75 - 217.1 = -49.35\,\text{mL}$ Convert to liters: $\Delta V = -0.04935\,\text{L}$

Therefore, $w = -(1\,\text{atm})(-0.04935\,\text{L}) = 0.04935\,\text{L atm}$ Using $1\,\text{L atm} = 101.325\,\text{J}$ we get $w = 0.04935 \times 101.325 \approx 5.0\,\text{J}$

4. Change in internal energy

By first law of thermodynamics, $\Delta U = q + w$ Thus, $\Delta U = -200 + 5 = -195\,\text{J}$

So, $\Delta U = -x\,\text{J} = -195\,\text{J}$ Hence, $x = 195$

5. Comparison with stored answer

Stored correct answer = $195$

This matches the derived answer.