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NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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What is entropy change in 2 mol N22 {~mol} {~N}_2, when its temperature is taken from 400 K400 {~K} to 800 K800 {~K}, adiabatically.
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Correct option: C
40JK40 \frac{{J}}{{K}}
The correct answer is 40 J/K. Since the process is adiabatic, q = 0. The entropy change is calculated using the formula ΔS=nCvln⁡(T2/T1)\Delta S = nC_v \ln(T_2/T_1) for an ideal gas.

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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
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