NEET · Chemistry —

Thermodynamics appeared 2 times across 1 year — 4.4% of Chemistry. This question is from First Law of Thermodynamics.

Year 2024 Total
Questions 2 2

At a certain temperature, T(K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:

Solution & Explanation

Related Formula
Δ U = q + w
Core Logic

Heat absorbed q = +500 J. Work done by system w = -200 J.

Δ U = 500 - 200 = 300 J
Step 1: Conclusion

Change in internal energy is 300 J.

Pattern Recognition

First Law sign convention: heat absorbed is positive, work done by system is negative.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Reference Study Guides

More Thermodynamics Previous-Year Questions

Q85 neet_2024_05_may_morning Gibbs Free Energy and Spontaneity
Consider the following reaction: 2A(g) + B(g) arrow 2D(g) Δ U° = -10 kJ mol⁻¹ and Δ S° = -44 J K⁻¹ at 298 K Identify the correct option with Δ G° for the reaction and spontaneity of the reaction at 298 K. (Given: R = 8.31 J mol⁻¹ K⁻¹)
  • A. (1) -1.635 kJ mol⁻¹, spontaneous
  • B. (2) +0.63568 kJ mol⁻¹, non-spontaneous
  • C. (3) -0.63568 kJ mol⁻¹, spontaneous
  • D. (4) +1.635 kJ mol⁻¹, non-spontaneous

Solution

Related Formula
Δ H° = Δ U° + Δ ng RT Δ G° = Δ H° - TΔ S°
Core Logic
Δ ng = 2 - (2 + 1) = -1 Δ H° = -10 - (1 × 298 × 8.31)/(1000) = -10 - 2.47638 = -12.47638 kJ/mol Δ G° = -12.47638 - (298 × (-44))/(1000) = -12.47638 + 13.112 = +0.63562 kJ/mol

Since Δ G° > 0, process is non-spontaneous.

Step 1: Conclusion

Δ G° = +0.63568 kJ mol⁻¹ and non-spontaneous.

Pattern Recognition

Positive Δ G° Non-spontaneous reaction under standard conditions.

Chapter Mix

Class 11 Chemistry: Thermodynamics

More Thermodynamics Questions — neet_2024_05_may_morning

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