A gas expands at constant pressure. The work done by the gas is given by: MCQ with Answer and Explanation

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A gas expands at constant pressure. The work done by the gas is given by:

A. VΔP

B. P/V

C. PΔV

D. Δ(PV)

Answer: Option C

Solution (By JKSSB Mock Tests)

For constant pressure (isobaric) process, work done by gas W = PΔV, where ΔV is change in volume. This derives from mechanical work W = F·d and pressure P = F/A, so F = PA, and dV = A·d, thus W = P·dV. Option D Δ(PV) applies to other contexts like enthalpy. Memory tip: 'Isobaric work = P times volume change'. This fundamental thermodynamics result is frequently used in first law applications. Competitive exams often combine this with ideal gas law for numerical problems; always verify process constraints (constant P here).

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The speed of sound in air at 0°C is approximately 332 m/s. At 27°C, it will be approximately:

A. 332 m/s

B. 340 m/s

C. 360 m/s

D. 350 m/s

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Correct Answer: Option D

Explanation:
Speed of sound in air v ∝ √T, where T is absolute temperature in Kelvin. T₀ = 0°C = 273 K, v₀ = 332 m/s. T = 27°C = 300 K. Thus v = v₀√(T/T₀) = 332 × √(300/273) ≈ 332 × √1.099 ≈ 332 × 1.048 ≈ 348 m/s ≈ 350 m/s. Approximate rule: speed increases by 0.6 m/s per °C rise, so 27×0.6≈16.2, 332+16.2≈348.2 m/s. Memory aid: 'v ∝ √T in Kelvin'. Such temperature-dependence problems test application of sound wave properties, common in competitive exams with emphasis on absolute temperature usage.

This question belongs to: Science Physics