NOTE** Charles Law problems must have the PRESSURE CONSTANT. Try using Charles’s law to solve the following problem. Show Example A sample of gas at 15ºC and 1 atm has a volume of 2.50 L. What volume will this gas occupy at 30ºC and 1 atm? OK, Here is our formula, . So, just plug in the numbers. NO As a Chemistry teacher I am REQUIRED to give you the temperature in Celsius and make you convert to Kelvin. It is the Law of all Chemistry Teachers. So convert the temperatures to Kelvin. T1= 15C +273=288K and T2 =30C +273=303K Now Plug in and Solve
This makes sense—the temperature is increasing slightly, so the volume should increase slightly. Again be careful of questions like this. It’s tempting to just use the Celsius temperature, but you must first convert to Kelvin temperature (by adding 273) to get the correct relationships!
In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation. 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
Charles Law Chemistry Questions with SolutionsQ1. Suppose P, V, and T denote the gas’s pressure, volume, and temperature. In that case, the correct representation of Chale’s law is
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?
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?
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?
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
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.
Answer:
Q11. Calculate the decrease in temperature (in Celsius) when 2.00 L at 21.0 °C is compressed to 1.00 L. Answer: Given 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? Answer: Given 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? Answer: Given 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? Answer: Given 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? Answer: Given 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 LawQ1. 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. Recommended VideosIdeal Gas Equation definitions, derivation
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