Class 9 Study of Gas Laws Mock Test (ICSE)

Topics in Study of Gas Laws

• Kinetic theory of gases
• Boyle's law
• Charles' law
• Gas equation
• Standard temperature and pressure

Study of Gas Laws practice questions with answers

1. Q1Multiple choice·1 mark
   Which of the following sets of conditions represents standard temperature and pressure (STP) as used in gas law calculations?

   1. (a)$273^\circ\text{C}$ and $760\text{ mm of Hg}$
   2. (b)$0^\circ\text{C}$ and $1\text{ mm of Hg}$
   3. (c)$0^\circ\text{C}$ and $760\text{ mm of Hg}$
   4. (d)$25^\circ\text{C}$ and $1\text{ atm}$
   Show answer and step marks

   1. Correct option: (c) $0^\circ\text{C}$ and $760\text{ mm of Hg}$. Standard temperature is $0^\circ\text{C}$ ($273\text{ K}$) and standard pressure is $760\text{ mm of Hg}$ (or $1\text{ atm}$).+1

   Answer: (c)
2. Q2Multiple choice·1 mark
   According to the kinetic theory of gases, collisions between gas molecules and the walls of the container are:

   1. (a)Perfectly elastic with no loss of total kinetic energy
   2. (b)Associated with a net gain in momentum of the system
   3. (c)Involving chemical reactions at high speeds
   4. (d)Perfectly inelastic losing kinetic energy
   Show answer and step marks

   1. Correct option: (a) Perfectly elastic with no loss of total kinetic energy.+1

   Answer: (a)
3. Q3Multiple choice·1 mark
   The combined gas equation relating pressure, volume, and absolute temperature for a fixed mass of a dry gas is expressed as:

   1. (a)$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
   2. (b)$\frac{P_1 T_1}{V_1} = \frac{P_2 T_2}{V_2}$
   3. (c)$\frac{T_1 V_1}{P_1} = \frac{T_2 V_2}{P_2}$
   4. (d)$P_1 V_1 T_1 = P_2 V_2 T_2$
   Show answer and step marks

   1. Correct option: (a) $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.+1

   Answer: (a)
4. Q4Short answer·2 marks
   State Boyle's law. A gas occupies $500\text{ cm}^3$ at a pressure of $750\text{ mm of Hg}$. Find its volume if the pressure is increased to $1250\text{ mm of Hg}$ at constant temperature.
   Show answer and step marks

   1. State Boyle's law: The volume of a given mass of a dry gas is inversely proportional to its pressure at constant temperature.+1
   2. Apply $P_1V_1 = P_2V_2$: $750 \times 500 = 1250 \times V_2 \implies V_2 = \frac{750 \times 500}{1250} = 300\text{ cm}^3$.+1

   Answer: Boyle's law: at constant temperature, the volume of a given mass of a dry gas is inversely proportional to its pressure. Volume at $1250\text{ mm of Hg}$ = $300\text{ cm}^3$.
5. Q5Long answer·4 marks
   (a) A sample of carbon dioxide gas occupies $400\text{ cm}^3$ at $27\text{ }^\circ\text{C}$ under constant pressure. Calculate its volume at $-3\text{ }^\circ\text{C}$ at the same pressure. (b) (i) State Charles' law. (ii) Why is it necessary to use the absolute temperature scale while solving problems based on Charles' law?
   Show answer and step marks

   1. Convert initial and final Celsius temperatures to Kelvin: $T_1 = 27 + 273 = 300\text{ K}$, $T_2 = -3 + 273 = 270\text{ K}$.+0.5
   2. Write Charles' law formula: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$+0.5
   3. Substitute the given values into the formula: $V_2 = \frac{400 \times 270}{300}$+0.5
   4. Calculate the final volume with proper unit: $360\text{ cm}^3$+0.5
   5. State Charles' law: The volume of a given mass of a dry gas is directly proportional to its absolute temperature if pressure remains constant.+1
   6. Explain necessity of absolute temperature scale: Celsius scale temperatures can be zero or negative, leading to physically impossible zero or negative volumes, whereas the Kelvin scale begins at absolute zero.+1

   Answer: (a) $360\text{ cm}^3$. (b)(i) At constant pressure, the volume of a given mass of a dry gas is directly proportional to its absolute (Kelvin) temperature. (ii) Celsius temperatures can be zero or negative, which would give zero or negative volumes; the Kelvin scale starts at absolute zero, so all temperatures on it are positive.
6. Q6Long answer·4 marks
   (a) State Charles' law. What is the value of absolute zero on the Celsius scale? (b) A given mass of a gas occupies $300\text{ cm}^3$ at $27\text{ }^\circ\text{C}$. Calculate the volume of the gas at $127\text{ }^\circ\text{C}$ at constant pressure.
   Show answer and step marks

   1. State Charles' law: At constant pressure, the volume of a given mass of a dry gas is directly proportional to its absolute temperature.+1
   2. Absolute zero on the Celsius scale: $-273\text{ }^\circ\text{C}$ (also credit $-273.15\text{ }^\circ\text{C}$).+1
   3. Convert temperatures to Kelvin: $T_1 = 27 + 273 = 300\text{ K}$ and $T_2 = 127 + 273 = 400\text{ K}$, and set up the relation $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.+1
   4. Substitute values: $\frac{300}{300} = \frac{V_2}{400}$, which gives $V_2 = 400\text{ cm}^3$.+1

   Answer: (a) At constant pressure, the volume of a given mass of a dry gas is directly proportional to its absolute temperature; absolute zero is $-273\text{ }^\circ\text{C}$. (b) $400\text{ cm}^3$

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