L Explanation: We're asked to find the volume occupied by 454
g of H 2 at 1.05
atm and 25
o C . To do this, we can use the ideal-gas equation: p V = n R T where p is the pressure, in units of atm (given as 1.05
atm )
V is the volume the gas occupies, in units of L (we'll be finding this)
n is the number of moles of the gas present (we'll need to convert the given mass in grams to moles)
R is the universal gas constant, equal to 0.082057 L ⋅ atm mol ⋅ K
T is the absolute temperature of the gas, in units of K (given 25
o C )
We need to do some conversions: Let's find the number of moles of H 2 present via the molar mass of H 2 ( 2.02
g/mol ): 454 g H 2 ( 1 l mol H 2 2.02 g H 2 ) = 225
mol H 2 The temperature in K is 25 o C + 273 = 298
K Plugging in known values, and solving for the volume, V , we have V = n R T p = ( 225 mol ) ( 0.082057 L ⋅ atm mol ⋅ K ) ( 298 K ) ( 1.05 atm ) = 5.24 × 10 3
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