N₂ gas behaves ideally at its Boyle temperature in the pressure range from 0 to 50 atm. The critical — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

Critical temp $T_c = 96 \text{ K}$. Boyle temp $T_B = \frac{27}{8} T_c = 3.375 \times 96 = 324 \text{ K} = 51^\circ\text{C}$. At $51^\circ\text{C}$ and $P \le 50 \text{ atm}$, it behaves ideally ($V_m = RT/P$). (a) At $T_c$, gas is non-ideal ($Z \approx 3/8$), so $V_m \neq 9.6 \text{ L}$. (b) At 127°C, gas is non-ideal, $V_m \neq 40 \text{ L}$ ideally calculated. (c) At 82.1 atm, pressure exceeds 50 atm threshold for ideal behavior at $T_B$. (d) At 51°C and 0.821 atm, it is ideal, so $V_m = 0.0821 \times 324 / 0.821 = 32.4 \text{ L}$, not 32.4 ml. All are mathematically incorrect. Therefore, correct answer is A,B,C,D.