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

Step-by-step 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 * V = 4.53 mol * 0.082 L atm K^(-1) mol^(-1) * 301 K`

V =
`(111.81 L atm)/(1.78 atm)`

= 62.81 L

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