Question

What is the pressure (in atm) of a 10.89 mol sample of helium gas at 19.3 degrees C if its volume is in 5.70 L?

Answer 1

To solve this problem, we can use the ideal gas law:

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 in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15:

T = 19.3 + 273.15 = 292.45 K

Next, we can plug in the given values and solve for P:

P = nRT/V

P = (10.89 mol)(0.08206 L·atm/mol·K)(292.45 K)/(5.70 L)

P = 47.4 atm

Therefore, the pressure of the helium gas is 47.4 atm.

Answer 2

the pressure of a 10.89 mol sample of helium gas at 19.3 degrees celsius and a volume of 5.70 liters is 1.92 atm (atmospheres). you can verify this number with the ideal gas equation (pV = nRT)