Thermal Properties Of Matter Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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?

Given: \( P_1 = 1 \, \text{atm} \), \( V_1 = 2 \, \text{L} \), \( T_1 = 300 \, \text{K} \), \( V_2 = 0.5 \, \text{L} \), \( T_2 = 450 \, \text{K} \).

\( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).

\( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 1 \times \frac{2}{0.5} \times \frac{450}{300} = 1 \times 4 \times 1.5 = 6 \, \text{atm} \).

5 atm
5.5 atm
6 atm
6.5 atm
3

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

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).

It expands more
It conducts heat poorly
It has low density
It requires more heat per unit temperature rise
4

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?

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).

20%
25%
15%
22%
1

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 = \( 3.35 \times 10^5 \, \text{J kg}^{-1} \))

\( Q_1 = m s \Delta T = 1 \times 2100 \times 10 = 21000 \, \text{J} \).

\( Q_2 = m L_f = 1 \times 3.35 \times 10^5 = 335000 \, \text{J} \).

Total heat: \( Q = Q_1 + Q_2 = 21000 + 335000 = 356000 \, \text{J} = 356 \, \text{kJ} \).

350 kJ
356 kJ
360 kJ
340 kJ
2

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?

Given: \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \), \( P_1 = 2 \, \text{atm} \), \( V_1 = 4 \, \text{L} \), \( T_2 = 127^\circ \text{C} = 400 \, \text{K} \), \( P_2 = 1 \, \text{atm} \).

Ideal gas law: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).

\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 4 \times \frac{2}{1} \times \frac{400}{300} = 4 \times 2 \times \frac{4}{3} = 10.67 \, \text{L} \).

10.67 L
10.5 L
11 L
12 L
1

How much heat is required to convert \( 0.2 \, \text{kg} \) of ice at \( -22^\circ \text{C} \) to steam at \( 110^\circ \text{C} \) in a \( 0.05 \, \text{kg} \) aluminium calorimeter initially at \( 30^\circ \text{C} \)? (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 = \( 2.256 \times 10^6 \, \text{J kg}^{-1} \), aluminium = \( 900 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( Q_1 = 0.2 \times 2100 \times 22 = 9240 \, \text{J} \) (ice to 0°C).

\( Q_2 = 0.2 \times 3.35 \times 10^5 = 67000 \, \text{J} \) (melting).

\( Q_3 = (0.2 \times 4186 + 0.05 \times 900) \times 100 = (837.2 + 45) \times 100 = 88220 \, \text{J} \) (to 100°C).

\( Q_4 = 0.2 \times 2.256 \times 10^6 = 451200 \, \text{J} \) (vaporization).

\( Q_5 = 0.2 \times 4186 \times 10 = 8372 \, \text{J} \) (steam to 110°C).

Calorimeter cools: \( Q_6 = 0.05 \times 900 \times (30 - 0) = 1350 \, \text{J} \) (assume it cools to 0°C).

Total: \( Q = 9240 + 67000 + 88220 + 451200 + 8372 - 1350 = 622682 \, \text{J} = 622.68 \, \text{kJ} \).

620 kJ
621 kJ
623 kJ
622.68 kJ
4

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 = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( Q_1 = 0.35 \times 2100 \times 28 = 20580 \, \text{J} \) (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: \( Q = 20580 + 117250 + 150370 + 789600 + 21976.5 - 965 = 1097811.5 \, \text{J} = 1097.81 \, \text{kJ} \).

1097 kJ
1098 kJ
1096 kJ
1097.81 kJ
4

A gas at \( 2 \, \text{atm} \) and \( 27^\circ \text{C} \) occupies \( 5 \, \text{L} \). What pressure is required to compress it to \( 2 \, \text{L} \) while maintaining the temperature at \( 327^\circ \text{C} \)?

Given: \( P_1 = 2 \, \text{atm} \), \( V_1 = 5 \, \text{L} \), \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \), \( V_2 = 2 \, \text{L} \), \( T_2 = 327^\circ \text{C} = 600 \, \text{K} \).

\( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).

\( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 2 \times \frac{5}{2} \times \frac{600}{300} = 2 \times 2.5 \times 2 = 10 \, \text{atm} \).

8 atm
9 atm
11 atm
10 atm
4

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?

Given: \( P_1 = 2.2 \, \text{atm} \), \( T_1 = 57^\circ \text{C} = 330 \, \text{K} \), \( V_1 = 5.5 \, \text{L} \), \( V_2 = 3.3 \, \text{L} \), \( T_2 = 107^\circ \text{C} = 380 \, \text{K} \).

\( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).

\( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 2.2 \times \frac{5.5}{3.3} \times \frac{380}{330} \).

\( P_2 = 2.2 \times 1.6667 \times 1.1515 \approx 4.22 \, \text{atm} \).

4.1 atm
4.2 atm
4.3 atm
4.22 atm
4

How much heat is required to vaporize \( 0.25 \, \text{kg} \) of ethanol at \( 78^\circ \text{C} \)? (Latent heat of vaporization of ethanol = \( 8.5 \times 10^5 \, \text{J kg}^{-1} \))

Given: \( m = 0.25 \, \text{kg} \), \( L_v = 8.5 \times 10^5 \, \text{J kg}^{-1} \).

\( Q = m L_v = 0.25 \times 8.5 \times 10^5 = 212500 \, \text{J} = 212.5 \, \text{kJ} \).

210 kJ
212.5 kJ
215 kJ
200 kJ
2

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