Question

What volume does 4.53 moles of hydrogen at 1.78 atm and 301 K occupy

Answer 1

V = nRT / P = (4.53 mol) x (0.08205746 L atm/K mol) x (301 K) / (1.78 atm) = 62.9 L

Answer 2

Explanation:

According to ideal gas equation, product of pressure and volume equals n times R times T.

Mathematically,            PV = nRT

where         P = pressure

                   V = volume

                   n = number of moles

                   R = gas constant

                   T = temperature

Since, it is given that no. of moles is 4.53 mol, pressure is 1.78 atm and temperature is 301 K. Value of R will be 0.082 $L atm K^{-1}mol^{-1}$ Therefore, calculate the volume as follows.

                                 PV = nRT

         $1.78 atm \times V = 4.53 mol \times 0.082 L atm K^{-1} mol^{-1} \times 301 K$

                              V = $\frac{111.81 L atm}{1.78 atm}$

                                  = 62.81 L

Thus, we can conclude that volume of hydrogen will be 62.81 L.