NEET Thermal Properties of Matter — Set 13

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Question 1

A gas at $ 1.8 \, \text{atm} $ and $ 27^\circ \text{C} $ occupies $ 3.5 \, \text{L} $. If the pressure is increased to $ 3 \, \text{atm} $ and temperature raised to $ 127^\circ \text{C} $, what is the percentage change in volume?

TopperSquare Expert · Accepted Answer

Given: $ P_1 = 1.8 \, \text{atm} $, $ T_1 = 27^\circ \text{C} = 300 \, \text{K} $, $ V_1 = 3.5 \, \text{L} $, $ P_2 = 3 \, \text{atm} $, $ T_2 = 127^\circ \text{C} = 400 \, \text{K} $. $ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} $. $ V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 3.5 \times \frac{1.8}{3} \times \frac{400}{300} = 3.5 \times 0.6 \times 1.3333 \approx 2.8 \, \text{L} $. Percentage change: $ \frac{V_1 - V_2}{V_1} \times 100 = \frac{3.5 - 2.8}{3.5} \times 100 = \frac{0.7}{3.5} \times 100 = 20\% $ (decrease).

Question 2

Why does a substance with high specific heat capacity take longer to heat up?

TopperSquare Expert · Accepted Answer

A high specific heat capacity means more heat ($ \Delta Q = m s \Delta T $) is required per unit mass to raise the temperature, slowing the heating process (Section 10.6).

Question 3

How much heat is required to convert $ 0.35 \, \text{kg} $ of ice at $ -28^\circ \text{C} $ to steam at $ 115^\circ \text{C} $ in a $ 0.1 \, \text{kg} $ brass calorimeter initially at $ 25^\circ \text{C} $? (Specific heat of ice = $ 2100 \, \text{J kg}^{-1} \text{K}^{-1} $, latent heat of fusion = $ 3.35 \times 10^5 \, \text{J kg}^{-1} $, specific heat of water = $ 4186 \, \text{J kg}^{-1} \text{K}^{-1} $, latent heat of vaporization = $ 2.256 \times 10^6 \, \text{J kg}^{-1} $, brass = $ 386 \, \text{J kg}^{-1} \text{K}^{-1} $)

TopperSquare Expert · Accepted Answer

$ Q_1 = 0.35 \times 2100 \times 28 = 20580 \, \text{J} $ (ice to 0°C). $ Q_2 = 0.35 \times 3.35 \times 10^5 = 117250 \, \text{J} $ (melting). $ Q_3 = (0.35 \times 4186 + 0.1 \times 386) \times 100 = (1465.1 + 38.6) \times 100 = 1503.7 \times 100 = 150370 \, \text{J} $ (to 100°C). $ Q_4 = 0.35 \times 2.256 \times 10^6 = 789600 \, \text{J} $ (vaporization). $ Q_5 = 0.35 \times 4186 \times 15 = 21976.5 \, \text{J} $ (steam to 115°C). Calorimeter cools: $ 0.1 \times 386 \times (25 - 0) = 965 \, \text{J} $. Total: $ Q = 20580 + 117250 + 150370 + 789600 + 21976.5 - 965 = 1097811.5 \, \text{J} = 1097.81 \, \text{kJ} $.

NEET Physics — Thermal Properties of Matter

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NEET Physics: Thermal Properties of Matter — MCQ Test 13

Q6. How much heat is required to convert \( 1 \, \text{kg} \) of ice at \( -10^\circ \text{C} \) to water at \( 0^\circ \text{C} \)? (Specific heat of ice = \( 2100 \, \text{J kg}^{-1} \text{K}^{-1} \), Latent heat of fusion = \( 3.35 \times 10^5 \, \text{J kg}^{-1} \))

Q7. A gas at \( 2.2 \, \text{atm} \) and \( 57^\circ \text{C} \) occupies \( 5.5 \, \text{L} \). If the volume is reduced to \( 3.3 \, \text{L} \) and temperature increased to \( 107^\circ \text{C} \), what is the final pressure?

Q9. A gas at \( 1 \, \text{atm} \) and \( 300 \, \text{K} \) occupies \( 2 \, \text{L} \). If it is compressed to \( 0.5 \, \text{L} \) while temperature rises to \( 450 \, \text{K} \), what is the final pressure?

Q10. A gas at \( 27^\circ \text{C} \) and \( 2 \, \text{atm} \) occupies \( 4 \, \text{L} \). If the temperature is increased to \( 127^\circ \text{C} \) and pressure is reduced to \( 1 \, \text{atm} \), what is the new volume?