Question

What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C?

Answer 1

`V=(nRT)/P`

Explanation:

`V=(0.8*molxx0.0821*L*atm*K^-1mol^(-1)*500*K)/(5.3*atm)`

`=` `??L`

Answer 2

`=6.19L`

Explanation:

By Ideal gas equation we know,

`PV=nRT`

Where

• Pressure `P=5.3atm`
• Volume `V=?L`
• Universal gas constant `R=0.082LatmK^-1mol^-1`
• Number moles of gas `n =0.8mol`
• The temperature of the gas `T=227+273=500K`

Putting inthe gas equation we get

`5.3V=0.8xx0.082xx500`

`V=(0.8xx0.082xx500)/5.3L=6.19L`