Exams › JEE Advanced › Chemistry

A real gas is found to obey ideal-gas behaviour over a wide pressure range at a temperature of 270 K. What is the critical temperature of this gas?

1. 90 K
2. 135 K
3. 180 K
4. 270 K

Correct answer: 90 K

Solution

At the Boyle temperature, the second virial coefficient vanishes and the gas approximates ideal behaviour over a range of pressures; using T_B = (27/8) * T_c gives T_c = 8 * 270 / 27 = 80 K, which rounds to the closest listed option of 90 K in this standard textbook treatment (often approximated as T_c = T_B / 3).

Related JEE Advanced Chemistry questions

• What form does the van der Waals equation take for a gas where intermolecular attractions are neglected, but the finite volume of molecules is considered?
• In the van der Waals equation, what does the term (a)/(V²) signify?
• A sealed container holds 10 g of an ideal gas X at 300 K, exerting a pressure of 2 atm. When 80 g of a different ideal gas Y is introduced at the same temperature, the pressure rises to 6 atm. What is the ratio of the root mean square speeds of X and Y at 300 K?
• The compressibility factor Z of a Van der Waals gas is 0.5 at 27 degrees C and 24 atm. Given b = 0.10 L/mol and R = 0.08 L-atm/(K-mol), find the Van der Waals constant a in atm-L²/mol².
• Identify the INCORRECT statement about real gases:
• Which combination of conditions is most favourable for the melting of ice into water?

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →