Answer: The volume of a balloon that can hold 313.0 g of helium gas is 2174 L
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 635.4 mmHg = 0.8360 atm (760mmHg=1 atm)
V = Volume of gas = ?
n = number of moles = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{313.0g}{4g/mol}=78.25mol[/tex]
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]10.00^0C=(10.00+273)K=283.00K[/tex]
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{78.25mol\times 0.0820 Latm/K mol\times 283.00K}{0.8360atm}=2174L[/tex]
Thus volume of a balloon that can hold 313.0 g of helium gas is 2174 L
