Calculate the pressure, in atmospheres, of 2.80 moles helium gas in a 18.0 liters container at 24 degrees celcius.
To calculate the pressure, you can use the ideal gas law, which states: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin. The Kelvin temperature scale starts at absolute zero (-273.15 degrees Celsius), so we add 273.15 to convert. In this case, the temperature would be 24 + 273.15 = 297.15 K.
Next, we can substitute the known values into the ideal gas law equation:
PV = nRT
P * 18.0 L = 2.80 mol * 0.0821 L * atm / (mol * K) * 297.15 K
Now, let's solve for P:
P = (2.80 mol * 0.0821 L * atm / (mol * K) * 297.15 K) / 18.0 L
P = 36.38 atm
Therefore, the pressure of 2.80 moles of helium gas in a 18.0 liters container at 24 degrees Celsius is approximately 36.38 atmospheres.
The ideal gas law is PV = nRT. You have n (2.80mol), V (18L), and T (24°C). R is the universal gas constant (which equals 0.0821 L* atm/mol * K).
Convert Celsius to Kelvin:
24°C + 273K = 297K
insert the numbers into the equation:
P(18L) = (2.80mol)(0.0821 L atm/mol * K)(297K)
isolate P:
P = (2.80mol)(0.0821 L atm/mol * K)(297K)/(18L)
P = 0.178 atm