Pressure

What pressure will be exerted by 10 mole hydrogen gas in an 8.5L cylinder at 10°C?

• Interactive popup. Assistance may be required. The temperature is not in Kelvin. What do I do?

  Remember to convert temperatures that are given to K before solving for the problem. The temperature conversions are as follows:

  Converting Celsius to Kelvin
  °C + 237 = K

  Converting Fahrenheit to Kelvin
  First convert Fahrenheit to Celsius
  °F → deg;C
  °C = (°F – 32) / (1.8)
  °C → K
  °C + 237 = K

  
• Interactive popup. Assistance may be required. Hint 1
  V = 8.5L
  
  n = 10 mol
  
  R = 8.31 numerator: L times kPa, denominator: mol times K L * kPa mol * K
  
  T = 10°C = 20 + 273 = 283K
  
  P =? kPa
• Interactive popup. Assistance may be required. Hint 2
  PV = nRT
  
  P = nRT ÷ V
  
  P = 10 * 8.314 * 283 ÷ 8.5
  
  P = 23528.62 ÷ 8.5
  
  
• Interactive popup. Assistance may be required. Check Your Answer
  2768 kPa

Temperature

To what temperature does a 0.25L cylinder containing 0.10 mol helium gas need to be cooled in order for the pressure to be 253.25kPa?

• Interactive popup. Assistance may be required. Hint 1
  P = 253.25kPa
  
  V = 250mL = 0.250L
  
  R = 8.31 numerator: L times kPa, denominator: mol times K L * kPa mol * K
  
  T = ? K
• Interactive popup. Assistance may be required. Hint 2
  PV = nRT
  
  PV ÷ nR = T
  
  T = PV ÷ nR
  
  T = 253.25 * 0.250 ÷ 0.10 * 8.314
  
  T = 63.3125 ÷ 0.8314
• Interactive popup. Assistance may be required. Check Your Answer
  76.15 K

Volume

A balloon company is testing out new cylinders that will give them the most return on their investment. In order to keep costs low, they need to keep the volume below 1L. What volume is needed to store 0.050 moles of helium gas at 202.6kPa and 298K?

• Interactive popup. Assistance may be required. Hint 1
  P = 202.6 kPa
  
  n = 0.050 mol
  
  T = 298 K
  
  R = 8.31 numerator: L times kPa, denominator: mol times K L * kPa mol * K
  
  V = ? L
• Interactive popup. Assistance may be required. Hint 2
  PV = nRT
  
  V = nRT ÷ P
  
  V = 0.050 x 8.314 x 298 ÷ 202.6
  
  V = 123.88 ÷ 202.6
• Interactive popup. Assistance may be required. Check Your Answer
  0.615 L

Moles

A student is studying the ideal gas law in class and has been given a simulation to do. In the simulation, a 50L cylinder is filled with argon gas to a pressure of 10130.0kPa at 30°C. To finish the investigation, the student needs to calculate the number of moles in the 50L cylinder. How many moles of argon are in the cylinder?

• Interactive popup. Assistance may be required. Hint 1
  P = 10130.0kPa
  
  V = 50L
  
  R = 8.31 numerator: L times kPa, denominator: mol times K L * kPa mol * K
  
  T = 30°C = 30 + 273 = 303K
  
  n = ? mol
  
  
• Interactive popup. Assistance may be required. Hint 2
  PV = nRT
  
  PV ÷ RT = n
  
  n = PV ÷ RT
  
  n = 10130.0 * 50 ÷ 8.314 * 303
  
  n = 506500 ÷ 2519.142
• Interactive popup. Assistance may be required. Check Your Answer
  201.1 mol