A 1.5 mol sample of radon gas has a volume of 21.0L at 33°C.
What is the pressure of the gas?
A) 37.7 atm B) 1.8°C C) 1.8K D) 1.8 atm

The correct answer is D) 1.8 atm.

Explanation:

We use the ideal gas formula (PV= nRT). We first convert the unit of temperature in degrees Celsius to Kelvin. We use the ideal gas constant R = 0.082 l atm / K mol. We solve P:

273K =0°C ---> 33°C= 273 + 33=306 K

PV= nRT > P =(nRT)/ V

P= ( 1,5 mol x 0.082 l atm / K mol x 306 K)/ 21,0 L

P= 1,79 atm

dynamic equilibrium is the answer.