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Sagot :
The volume occupied by 22. 0 g of helium gas at 26. 0 ° C and 1. 20 atm of pressure is 112.37 L .
Calculation ,
According to ideal gas equation ,
PV = nRT ....(i)
where P is the pressure = 1. 20 atm
V is the volume of the helium gas = ?
R is the universal gas constant = 0.082 atm L /K mol
T is the temperature of the gas = 26. 0 ° C = 299 K
n is the number of moles
Number of miles (n) = given mass/ molar mass =22 g/4 = 5.5 moles
By putting the value of pressure , volume , temperature and universal gas constant in equation ( i) we get
1. 20 atm ×V = 5.5 moles × 0.082×299
V = 5.5 moles × 0.082×299/1. 20 atm = 112.37 L
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