Thermal Properties of Matter Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A gas at \( 67^\circ \text{C} \) and \( 1.6 \, \text{atm} \) occupies \( 8 \, \text{L} \). If the temperature is reduced to \( 17^\circ \text{C} \) and volume adjusted to \( 6 \, \text{L} \), what is the final pressure?

Given: \( T_1 = 67^\circ \text{C} = 340 \, \text{K} \), \( P_1 = 1.6 \, \text{atm} \), \( V_1 = 8 \, \text{L} \), \( T_2 = 17^\circ \text{C} = 290 \, \text{K} \), \( V_2 = 6 \, \text{L} \).

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

\( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 1.6 \times \frac{8}{6} \times \frac{290}{340} \).

\( P_2 = 1.6 \times 1.3333 \times 0.8529 \approx 1.82 \, \text{atm} \).

1.82 atm
1.8 atm
1.9 atm
1.7 atm
1

Which statement is true about the ideal gas equation?

The ideal gas equation (\( PV = nRT \)) relates pressure, volume, and temperature for an ideal gas, where \( R \) is the universal gas constant and \( n \) is the number of moles.

It applies only to real gases
It ignores temperature
It relates pressure and volume only
It relates pressure, volume, and temperature
4

What process occurs when dry ice changes directly from solid to vapor?

Sublimation is the direct transition from solid to vapor without passing through the liquid state, as seen with dry ice (Section 10.8).

Melting
Vaporization
Fusion
Sublimation
4

A gas at \( 2 \, \text{atm} \) and \( 47^\circ \text{C} \) occupies \( 6 \, \text{L} \). If the temperature drops to \( -3^\circ \text{C} \) and pressure becomes \( 1.5 \, \text{atm} \), what is the new volume?

Given: \( P_1 = 2 \, \text{atm} \), \( T_1 = 47^\circ \text{C} = 320 \, \text{K} \), \( V_1 = 6 \, \text{L} \), \( T_2 = -3^\circ \text{C} = 270 \, \text{K} \), \( P_2 = 1.5 \, \text{atm} \).

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

\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 6 \times \frac{2}{1.5} \times \frac{270}{320} = 6 \times 1.3333 \times 0.84375 \approx 6.75 \, \text{L} \).

6.7 L
6.75 L
6.8 L
6.9 L
2

Why does a metal rod expand more than a glass rod for the same temperature increase?

Metals have a higher coefficient of linear expansion (\( \alpha_l \)) than glass (Section 10.5, Table 10.1). For the same \( \Delta T \), \( \Delta L = L_0 \alpha_l \Delta T \) is greater for metals due to their larger \( \alpha_l \).

Metals have a higher coefficient of linear expansion
Glass has a higher specific heat capacity
Metals conduct heat better
Glass contracts on heating
1

Why does steam at \( 100^\circ \text{C} \) cause more severe burns than water at \( 100^\circ \text{C} \)?

Steam releases additional latent heat of vaporization when it condenses on skin, transferring more energy than liquid water at the same temperature.

Steam has higher pressure
Water cools faster
Steam has lower density
Steam releases latent heat on condensing
4

How much heat is required to convert \( 0.25 \, \text{kg} \) of ice at \( -15^\circ \text{C} \) to water at \( 60^\circ \text{C} \) in a \( 0.15 \, \text{kg} \) iron calorimeter initially at \( 20^\circ \text{C} \)? (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 = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), iron = \( 450 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( Q_1 = 0.25 \times 2100 \times 15 = 7875 \, \text{J} \) (ice to 0°C).

\( Q_2 = 0.25 \times 3.35 \times 10^5 = 83750 \, \text{J} \) (melting).

Heat gained: \( (0.25 \times 4186 + 0.15 \times 450) \times (60 - 0) = (1046.5 + 67.5) \times 60 = 1114 \times 60 = 66840 \, \text{J} \).

Calorimeter cools: \( 0.15 \times 450 \times (20 - 0) = 1350 \, \text{J} \).

Total: \( Q = 7875 + 83750 + 66840 - 1350 = 157115 \, \text{J} = 157.12 \, \text{kJ} \).

157 kJ
157.12 kJ
158 kJ
156 kJ
2

A gas occupies \( 5 \, \text{L} \) at \( 2 \, \text{atm} \) and \( 27^\circ \text{C} \). If the pressure is increased to \( 4 \, \text{atm} \) and volume reduced to \( 3 \, \text{L} \), what is the final temperature in Celsius?

Given: \( V_1 = 5 \, \text{L} \), \( P_1 = 2 \, \text{atm} \), \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \), \( P_2 = 4 \, \text{atm} \), \( V_2 = 3 \, \text{L} \).

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

\( T_2 = T_1 \times \frac{P_2}{P_1} \times \frac{V_2}{V_1} = 300 \times \frac{4}{2} \times \frac{3}{5} = 300 \times 2 \times 0.6 = 360 \, \text{K} \).

\( t_C = T_2 - 273.15 = 360 - 273.15 \approx 86.85^\circ \text{C} \approx 87^\circ \text{C} \).

87°C
85°C
90°C
88°C
1

A gas at \( 1 \, \text{atm} \) and \( 27^\circ \text{C} \) occupies \( 5 \, \text{L} \). What will be its volume at \( 2 \, \text{atm} \) and \( 127^\circ \text{C} \)?

Given: \( P_1 = 1 \, \text{atm} \), \( V_1 = 5 \, \text{L} \), \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \), \( P_2 = 2 \, \text{atm} \), \( T_2 = 127^\circ \text{C} = 400 \, \text{K} \).

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

\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 5 \times \frac{1}{2} \times \frac{400}{300} = 5 \times 0.5 \times 1.33 \approx 3.33 \, \text{L} \).

3.0 L
3.33 L
3.5 L
4.0 L
2

Why does a calorimeter use a material with low specific heat capacity for its container?

A material with low specific heat capacity absorbs less heat, minimizing its effect on the heat exchange between the substances being measured, ensuring accurate results.

To increase heat loss
To expand more
To absorb less heat
To conduct heat better
3

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