A sealed bottle contains air at 22 ° C and a pressure 12 . 0 × 10 5 Pa . The temperature is raised to 120 . 0 ° C . Calculate the new pressure.

Given, Temperature, T 1 = 22 ° C = 22 + 273 . 15 = 295 . 15 K , Pressure, P 1 = 12 . 0 × 10 5 Pa T 2 = 120 . 0 ° C = 120 + 273 . 15 = 393 . 15 K , P 2 = ? From Gay- Lussac's law, the pressure is directly proportional to the temperature at the constant volume. P 2 P 1 = T 2 T 1 P 2 = P 1 T 2 T 1 P 2 = 12 . 0 × 10 5 × 393 . 15 295 . 15 = 15 . 98 × 10 5 Pa Hence, the new pressure is 15 . 98 × 10 5 Pa .

A gas has pressure 8 . 2 × 10 6 Pa and volume 2 . 3 × 10 - 3 m 3 . The pressure is reduced to 4 . 5 × 10 6 Pa at constant temperature. Calculate the new volume of the gas.

Given, P 1 = 8 . 2 × 10 6 Pa , V 1 = 2 . 3 × 10 - 3 m 3 P 2 = 4 . 5 × 10 6 Pa , V 2 = ? The pressure is inversely proportional to the volume at the constant temperature. V 2 V 1 = P 1 P 2 V 2 = P 1 P 2 V 1 V 2 = 8 . 2 × 10 6 4 . 5 × 10 6 × 2 . 3 × 10 - 3 V 2 = 4 . 1911 × 10 - 3 m - 3 Hence, the new volume of the gas is 4 . 1911 × 10 - 3 m - 3 .

A mass of 12 . 0 kg of helium is required to fill a bottle with volume 5 . 00 × 10 - 3 m 3 at a temperature of 20 . 0 ° C . Determine the pressure in helium.

Given, The given mass of Helium, m = 12 . 0 × 10 3 g The volume required to fill a bottle, V = 5 . 00 × 10 - 3 m 3 Temperature, T = 20 ° C = 20 + 273 . 15 = 293 . 15 K The molecular mass of the Helium, M = 4 g / mol We know that, n = m M n = 12 . 0 × 10 3 4 = 3000 mol From the ideal gas equation, P V = n R T P = n R T V P = 3000 × 8 . 31 × 293 . 15 5 . 00 × 10 - 3 P = 1 . 461 × 10 9 Pa Hence, the pressure in helium is 1 . 461 × 10 9 Pa .

Determine the mass of carbon dioxide required to fill a tank with volume 12 . 0 × 10 - 3 m 3 at a temperature of 20 . 0 ° C and a pressure of 4 . 00 atm .

Given, The molecular mass of carbon dioxide, M = 44 g / mol The volume required to fill a tank, V = 12 . 0 × 10 - 3 m 3 Temperature, T = 20 . 0 ° C = 20 . 0 + 273 = 293 K Pressure, P = 4 . 00 atm = 4 × 1 . 013 × 10 5 = 4 . 052 × 10 5 Pa From the ideal gas equation, P V = m M R T P V M = m R T m = P V M R T m = 4 . 052 × 10 5 × 12 . 0 × 10 - 3 × 44 8 . 31 × 293 m = 87 . 86 g Hence, the mass of the carbon dioxide is 87 . 86 g .