Charles\' Law states that the volume of a gas varies directly with the Kelvin te
ID: 1027491 • Letter: C
Question
Charles' Law states that the volume of a gas varies directly with the Kelvin temperature. assuming that pressure is constant. We use the following formulas K = "C + 273 Solve the following problems assuming a constant pressure A sample of nitrogen occupies a volume of 250 mL at 25° C. occupy at 95° C? 1. What volume will it 2. Oxygen gas is at a temperature of 40° C when it occupies a volume of 2.3 liters. To what temperature should it be raised to occupy a volume of 6.5 liters? Hydrogen gas was cooled from 150 C to 50P C. Its new volume is 75 mL. What was its original volume? 3. Chlorine gas occuples a volume of 25 ml at 300 K. What volume will it occupy at 600 K? 4. A sample of neon gas at 50° C and a volume of 2.5 liters is cooled to 25 C. What is the new volume? 5. Fluorine gas at 3o5 K occupies a volume of 500mL. To what temperature should it be lowered to bring the volume to 300 mL? 6. 7. Helium occupies a volume of 3.8 liters at -45 C. What volume will it occupy at 45 C A sample of argon gas is cooled and its volume went from 380 mL to 250 mL. If its final temperature was-55° C, what was its original temperature? 8.Explanation / Answer
1) According to charles law V1/T1 = V2/T2, Here Temperature is converted onto kelvin so T1 = 25+273 = 298 K
so T1 = 298 K , V1 = 250 ml , T2 = 95+273 = 368 K
so V2 = V1 x T2 / T1
= 250 x 368 / 298 = 308.7 ml .
2) T1 = 40 + 273 = 313 K
V1 = 2.3 L = 2300 ml ,
V2 = 6.5 L = 6500 ml
T2 = V2 x T1/ V1
T2 = 6500 x 313 / 2300
= 884.6 K
converting in degree celcius = 884.6 - 273 = 611.60 C
T2 = 611.60 C
3) T1 = 423 K
T2 = 323 K
V2 = 75 ml
V1 = V2 x T1 / T2
= 75 x 423 / 323
= 98.2 ml
4) V1 = 25 ml
T1 = 300 K
T2 = 600 K
V2 = V1 x T2 / T1
= 25 x 600 / 300
= 50 ml
5) V1 = 2.5 ml
T1 = 50 + 273 =323 K
T2 = 25 + 273 = 298 K
V2 = V1 x T2 / T1
= 2.5 x 298 / 323
= 2.3 L