AP Physics Practice Questions: Thermodynamics

AP Physics multiple choice questions on thermodynamics, heat transfer, and ideal gases. Each question includes detailed solutions at the bottom of the page.

  1. An ideal gas at 300 K occupies 2.0 L at 1.0 atm. If the gas is compressed to 1.0 L at constant temperature, what is the new pressure?

    A) 0.5 atm
    B) 1.0 atm
    C) 2.0 atm
    D) 3.0 atm
    E) 4.0 atm

  2. How much heat is required to raise the temperature of 0.50 kg of water from 20°C to 80°C? (specific heat of water = 4186 J/kg·°C)

    A) 62.8 kJ
    B) 125.6 kJ
    C) 188.4 kJ
    D) 251.2 kJ
    E) 314.0 kJ

  3. A heat engine absorbs 500 J from a hot reservoir at 400 K and rejects 300 J to a cold reservoir. What is its maximum possible efficiency?

    A) 20%
    B) 40%
    C) 60%
    D) 80%
    E) 100%

  4. An aluminum rod (α = 23×10⁻⁶ /°C) is 2.0 m long at 20°C. How much does it expand when heated to 100°C?

    A) 0.37 mm
    B) 0.74 mm
    C) 1.47 mm
    D) 2.94 mm
    E) 3.68 mm

  5. 0.10 moles of an ideal monatomic gas undergo an isobaric expansion from 1.0 L to 3.0 L at 2.0 atm. How much work is done by the gas?

    A) 202.6 J
    B) 303.9 J
    C) 405.2 J
    D) 506.5 J
    E) 607.8 J

Solutions

  1. C) 2.0 atm
    Boyle's law: \( P_1 V_1 = P_2 V_2 \)
    \( 1.0 \times 2.0 = P_2 \times 1.0 \)
    \( P_2 = 2.0 \, \text{atm} \)

  2. B) 125.6 kJ
    \( Q = mc\Delta T = 0.50 \times 4186 \times 60 = 0.50 \times 251,160 = 125,580 \, \text{J} = 125.6 \, \text{kJ} \)

  3. B) 40%
    Actual efficiency = \( \frac{W}{Q_h} = \frac{Q_h - Q_c}{Q_h} = \frac{500 - 300}{500} = \frac{200}{500} = 0.4 = 40\% \)
    Carnot efficiency = \( 1 - \frac{T_c}{T_h} \), but we don't know T_c.

  4. E) 3.68 mm
    \( \Delta L = \alpha L_0 \Delta T = 23 \times 10^{-6} \times 2.0 \times 80 = 3.68 \times 10^{-3} \, \text{m} = 3.68 \, \text{mm} \)

  5. C) 405.2 J
    Work in isobaric process: \( W = P\Delta V \)
    \( P = 2.0 \, \text{atm} = 2.0 \times 101,325 = 202,650 \, \text{Pa} \)
    \( \Delta V = 3.0 - 1.0 = 2.0 \, \text{L} = 0.002 \, \text{m}^3 \)
    \( W = 202,650 \times 0.002 = 405.3 \, \text{J} \)