Considering entropy (S) as a thermodynamic parameter, the criterion

Considering Entropy S As A Thermodynamic Parameter The Criterion For Chemistry Question

Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is

Options

(a) ΔS system + ΔS surroundings > 0
(b) ΔS system – ΔS surroundings > 0
(c) ΔS system > 0 only
(d) ΔS surroundings > 0 only

Correct Answer:

ΔS system + ΔS surroundings > 0

Explanation:

ΔS = Rln V₂/V₁,
Here the volume of gas increases from V₁ to V₂ at consant temperature T. The total increase in entropy of the system and its surroundings during the spontaneous process of expansion considered above is, thus R ln (V₂/V₁).Since V₂ > V₁, it is obvious that the spontaneous (irreversible) isothermal expansion of a gas is accompanied by a increase in the entropy of the system and its surroundings considered together.
ΔS(sys) + ΔS(surr > 0.

Related Questions:

  1. Magnesium carbonate of x g decomposes. What is the percentage of purity
  2. A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength
  3. Number of chiral carbon in β-D-(+) glucose is
  4. When 0.1 mol MnO₄²⁻ is oxidised the quantity of electricity required to completely
  5. Acetylation of a secondary amine in alkaline medium yields

Topics: Thermodynamics (179)
Subject: Chemistry (2512)

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*