Approximately 0.271 grams of helium is necessary to reduce the pressure to 77 atm, given an initial volume of 334 mL, pressure of 5.00 atm, and temperature of 23°C.
To calculate the grams of helium that must be released to reduce the pressure to 77 atm, we can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. First, we need to determine the initial number of moles of helium using the given volume, pressure, and temperature. Then, we can use the molar mass of helium to convert moles to grams.
Given:
Volume (V) = 334 mL = 0.334 L
Pressure (P) = 5.00 atm
Temperature (T) = 23°C = 296 K
Molar mass of helium (M) = 4.0026 g/mol
Using the ideal gas law, we can solve for the number of moles of helium:
n = PV/RT = (5.00 atm) * (0.334 L) / (0.0821 L·atm/mol·K) * (296 K) ≈ 0.0676 moles
To find the mass of this amount of helium, we can use the molar mass:
Mass = n * M = 0.0676 moles * 4.0026 g/mol ≈ 0.271 g
Therefore, approximately 0.271 grams of helium must be released to reduce the pressure to 77 atm assuming ideal gas behavior.
The question probable may be:
How many grams of helium must be released to reduce the pressure to 77 atm assuming ideal gas behavior? Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of helium at 23°C.