 # Charles Law Questions

Charle’s law is also referred to as the law of volumes. It tells us about the behaviour of gases. Charle’s law states that the volume is directly proportional to the temperature of the gas at constant pressure.

V ∝ T

Or, V / T = k

Or, V = kT

 Definition: Charle’s law states that at constant pressure, the volume of a gas is directly proportional to the temperature.

## Charles Law Chemistry Questions with Solutions

Q1. Suppose P, V, and T denote the gas’s pressure, volume, and temperature. In that case, the correct representation of Chale’s law is

1. V is directly proportional to T (at constant P)
2. V inversely proportional to T (at constant P)
3. PV = nRT
4. None of the above

Answer: (a), If P, V, and T denote the gas’s pressure, volume, and temperature, then the correct representation of Chale’s law is V is directly proportional to T (at constant P).

V ∝ T

Or, V / T = k

Or, V = kT

Q2. How can we convert a Celsius temperature to Kelvin temperature?

2. By subtracting 37
3. By subtracting 273

Answer: (d), We can convert a Celsius temperature to Kelvin temperature by adding 273 to it.

Q3. Which element should remain constant if Charle’s law is applied to a gas sample?

1. Temperature and the number of moles of a gas
2. Pressure and the number of moles of a gas
3. Volume and the number of moles of a gas
4. Pressure only

Answer: (d), If Charle’s law is applied than the pressure of the gas sample should remain constant.

Q4. What is the value of the gas constant R?

1. 8.314 J mol-1 K-1
2. 0.082 J litre atm
3. 0.987 cal mol-1 K-1
4. 83 erg mol-1 K-1

Answer: (a), The value of gas constant R is 8.314 J mol-1 K-1.

Q5. According to Charle’s law, if the temperature of a gas at constant pressure is increased, the volume will also

1. Increase
2. Decrease
3. Remains the same
4. Can’t be determined

Answer: (a), Charle’s law states that volume is directly proportional to the temperature at constant pressure.

V ∝ T

Or, V / T = k

Or, V = kT.

Thus, if the system’s temperature increases, its volume will increase.

Q6. What is Charle’s law?

Answer: Charle’s law states that the volume of the gas is directly proportional to the absolute temperature of the gas at constant pressure.

V ∝ T

Or, V / T = k

Or, V = kT.

Thus, if the system’s temperature increases, its volume will increase or if the system’s temperature decreases, its volume will decrease.

Q7. Do you encounter any of the applications of Charle’s law in everyday life? If yes, Where?

Answer: Yes, we encounter applications of Charle’s law in everyday life. When we take a volleyball outside on a hot day, the ball expands a bit. As the temperature increases, its volume also increases, leading to the expansion of volleyball. Similarly, the volleyball shrinks on a cold day as the temperature drops; its size also decreases.

Q8. Is Charles Law indirect or direct relation?

Answer: Charle’s law is a direct relation between the temperature and the volume of the gas. When the molecule’s temperature rises, molecules move faster thereby creating more pressure on the gas container. Hence, increasing the volume of the container. If the pressure of the gas container remains constant then the number of the molecules also remains constant.

Q9. Can we use quantities in °C in Charle’s law?

Answer: No, we can not use quantities in °C in Charle’s law. The relationship between volume and temperature will work when the temperature is taken in kelvin while we can use any quantity for the volume of the gas.

Q10. Match the following.

Column 1

Column 2

Ideal Gas law

PTOTAL = P1 + P2 + P3 + P4 + . . . P(at constant volume and temperature)

Boyle’s law

V = kN (at constant pressure and temperature)

Charles’ law

V = kT (at constant pressure)

PV = k (at constant temperature)

Dalton’s law

PV = nRT

Column 1

Column 2

Ideal Gas law

PV = nRT

Boyle’s law

PV = k (at constant temperature)

Charles’ law

V = kT (at constant pressure)

V = kN (at constant pressure and temperature)

Dalton’s law

PTOTAL = P1 + P2 + P3 + P4 + . . . P(at constant volume and temperature)

Q11. Calculate the decrease in temperature (in Celsius) when 2.00 L at 21.0 °C is compressed to 1.00 L.

Initial Volume (V1 ) = 2 L

Initial Temperature (T1 ) = 21.0 °C = (21 + 273) K = 294 K

Final Volume (V2 ) = 1 L

To Find: Final Temperature (T2 ) = ?

We can calculate the final temperature of the gas using Charle’s law.

V1 / T1 = V2 / T2

2 / 294 = 1 / T2

T2 = 294 / 2

T2 = 147 K

T2 = (147 – 273) = – 126 °C

Hence the final temperature of the gas at volume 1 L is equivalent to – 126 °C.

Q12. A gas occupies a volume of 600.0 mL at a temperature of 20.0 °C. What will be its volume at 60.0 °C?

Initial Volume (V1 ) = 600.0 mL

Initial Temperature (T1 ) = 20.0 °C = (20 + 273) K = 293 K

Final Temperature (T2 ) = 60.0 °C = (60 + 273) K = 333 K

To Find: Final Volume (V2 ) = ?

We can calculate the final volume of the gas using Charle’s law.

V1 / T1 = V2 / T2

600 / 293 = V2 / 333

V2 = (600 X 333) / 293

V2 = 199800 / 293

V2 = 681.91 ≈ 682 mL

Hence the final volume of the gas at 60.0 °C is equivalent to 681.91 ≈ 682 mL.

Q13. A gas occupies a volume of 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0 °C?

Initial Volume (V1 ) = 900.0 mL

Initial Temperature (T1 ) = 27.0 °C = (27 + 273) K = 300 K

Final Temperature (T2 ) = 132.0 °C = (132 + 273) K = 405 K

To Find: Final Volume (V2 ) = ?

We can calculate the final volume of the gas using Charle’s law.

V1 / T1 = V2 / T2

900 / 300 = V2 / 405

V2 = (900 X 405) / 300

V2 = 364500 / 300

V2 = 1215 mL

Hence the final volume of the gas at 132.0 °C is equivalent to 1215 mL or 1.215 L.

Q14. What change in volume results if 60.0 mL of gas is cooled from 33.0 °C to 5.00 °C?

Initial Volume (V1 ) = 60.0 mL

Initial Temperature (T1 ) = 33.0 °C = ( 33 + 273) K = 306 K

Final Temperature (T2 ) = 5.0 °C = ( 5 + 273) K = 278 K

To Find: Final Volume (V2 ) = ?

We can calculate the final volume of the gas using Charle’s law.

V1 / T1 = V2 / T2

60 / 306 = V2 / 278

V2 = ( 60 X 278) / 306

V2 = 16680 / 306

V2 = 54.50 mL

Change in the volume = 60.0 – 54.5 = 5.5 mL.

Hence the change in the volume of the gas at 5.0 °C is equivalent to 5.5 mL.

Q15. A gas occupies a volume of 300.0 mL at a temperature of 17.0 °C. What is the volume at 10.0 °C?

Initial Volume (V1 ) = 300.0 mL

Initial Temperature (T1 ) = 17.0 °C = (17 + 273) K = 290 K

Final Temperature (T2 ) = 10.0 °C = ( 10 + 273) K = 283 K

To Find: Final Volume (V2 ) = ?

We can calculate the final volume of the gas using Charle’s law.

V1 / T1 = V2 / T2

300 / 290 = V2 / 283

V2 = (300 X 283) / 290

V2 = 84900 / 290

V2 = 292.75 mL

Hence the final volume of the gas at 10.0 °C is equivalent to 292.75 mL.

## Practise Questions on Charles Law

Q1. Differentiate between Boyle’s law and Charle’s law.

Q2. A gas occupies a volume of 500.0 mL at a temperature of 10.0 °C. What will be its volume at 50.0 °C?

Q3. A gas occupies a volume of 100.0 mL at a temperature of 27.0 °C. What is the volume at 10.0 °C?

Q4. What change in volume results if 10.0 mL of gas is cooled from 33.0 °C to 15.0 °C?

Q5. A gas occupies a volume of 1 L at a temperature of 17.0 °C. What is the volume at 10.0 °C?

Click the PDF to check the answers for Practice Questions.

## Ideal Gas Equation definitions, derivation 