Answer: 204 L
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 0.262 atm
V = Volume of gas = ?
n = number of moles = [tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{11.7g}{4g/mol}=2.92[/tex]
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]-50^0C=(273+(-50))K=223K[/tex]
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{2.92\times 0.0820 L atm/K mol\times 223K}{0.262atm}=204L[/tex]
Thus volume occupied by 11.7 g of Helium in a balloon is 204L
