Skip to main content
Back to THERMODYNAMICS

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Easy
1 of 1
Application +4 | -1
Heat exchanged during the boiling of water is 0.900 KJ/mol−10.900 \mathrm{~KJ} / \mathrm{mol}^{-1}. What will be the increase in entropy.
View answer and explanation
Correct option: A
2.41 JK−1 mol−12.41 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}
Option A (2.41 JK−1 mol−12.41 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}) is correct because it is calculated using the formula ΔS=qT\Delta S = \frac{q}{T} where T is the boiling point of water (373 K).

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →
The internal energy of one mole of a gas is: An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti{T_i} is the initial temperature and Tf{T_f} is the final temperature, which of the following statements is correct? For which of the process, ∆S is negative? The work done by a weightless piston in causing an expansion ∆V (at constant temperature), when the opposing pressure, P is variable, is given by: Assertion :- At constant pressure of the change H2O(s)→H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) work done is negative.

Reason :- During phase transition work done is always negative. For the reaction, X2O4(l)⟶2XO2(g),ΔU=2.1 kcal,ΔS=20 cal K−1\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g}), \Delta U=2.1 \mathrm{~kcal}, \Delta S=20 \mathrm{~cal} \mathrm{~K}^{-1} at 300 K300 \mathrm{~K}. Hence, ΔG\Delta G is
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →