Daily Olympiad: Physics - Thermal Physics [20260520]

Challenge yourself with today's KVPY practice! This test covers 'Thermal Physics' for Physics (KVPY - 12). Level: Hard | Duration: 45 mins.

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1. A gas undergoes an adiabatic expansion with γ = 5/3. How does the slope of the PV curve at any point compare to an isothermal process?

A. Slope is steeper for adiabatic B. Slope is steeper for isothermal C. Absolute magnitude depends on initial pressure D. Slope is same in magnitude for both

Solution

Correct: A

For an isothermal process, dP/dV = -P/V. For adiabatic, dP/dV = -γP/V. Since γ > 1, the adiabatic slope is steeper in magnitude. This requires differentiating both PV = nRT and PV^γ = constant.

2. A 5 kg block of ice at 0°C is melted into water and heated to 100°C. What is the total entropy change? (Latent heat of fusion = 344 kJ/kg, specific heat = 4.184 kJ/kg°C)

A. 2480 J/K B. 4560 J/K C. 6520 J/K D. 8230 J/K

Solution

Correct: D

Entropy change has two parts: melting and heating. ΔS = (mL)/T_fusion + m c ln(T2/T1) = (5×344000)/273 + 5×4184×ln(373/273) ≈ 6289 + 1873 ≈ 8162 J/K. Closest to 8230 J/K.

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Daily Olympiad: Physics - Thermal Physics [20260520]

1. A gas undergoes an adiabatic expansion with γ = 5/3. How does the slope of the PV curve at any point compare to an isothermal process?

2. A 5 kg block of ice at 0°C is melted into water and heated to 100°C. What is the total entropy change? (Latent heat of fusion = 344 kJ/kg, specific heat = 4.184 kJ/kg°C)

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