Thermodynamics

JEE Main 2024 (Online) 29th January Evening ShiftNumericalEasy📈 Topic in 17% papers+4

Ques. 1JEE Main 2024, 29 January (Evening Shift)

Standard enthalpy of vapourisation for CCl4 is 30.5 kJ mol-1. Heat required for vapourisation of 284 g of CCl4 at constant temperature is kJ. (Given molar mass in g mol-1; C = 12, Cl = 35.5)

Answer

Solution

Molar mass of CCl4 = 12 + 4(35.5) = 154 g mol-1

⇒ n = 284154 = 1.844 mol

∴ q = n × vapH = 1.844 × 30.5 = 56.2 kJ ≈ 56 kJ

Answer: 56

Keep practising — Thermodynamics

Ques. 2JEE Main 2024, 27 January (Morning Shift)

If three moles of an ideal gas at 300 K expand isothermally from 30 dm3 to 45 dm3 against a constant opposing pressure of 80 kPa, then the amount of heat transferred is J.

Ques. 3JEE Main 2024, 27 January (Evening Shift)

For a certain thermochemical reaction M N at T = 400 K, = 77.2 kJ mol-1, = 122 J K-1, log equilibrium constant (log K) is − 10-1.

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