Exams › JEE Advanced › Chemistry

For an isothermal expansion of an ideal gas, which combination of thermodynamic parameters is correct?

1. Delta_U = 0, Q = 0, w != 0 and Delta_H != 0
2. Delta_U != 0, Q != 0, w != 0 and Delta_H = 0
3. Delta_U = 0, Q != 0, w = 0 and Delta_H != 0
4. Delta_U = 0, Q != 0, w != 0 and Delta_H = 0

Correct answer: Delta_U = 0, Q != 0, w != 0 and Delta_H = 0

Solution

In isothermal expansion of an ideal gas: temperature is constant, so Delta_U = 0 and Delta_H = 0. Work is done by the gas (w != 0), and by first law Delta_U = Q + w (or Q - w depending on convention), so Q != 0.

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 →