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Sagot :
1) List known values
Mass: 32g (C4H10)
Temperature: 45.0ºC
Pressure: 728 mmHg
Ideal constant: 0.082057 L⋅atm⋅K−1⋅mol−1
List unknown values
Volume:
2) Convert grams of C4H10 into moles of C4H10
The molar mass ops C4H10 is 58.1222 g/mol
[tex]\text{molesofC}_4H_{10}=32gofC_4H_{10}\cdot\frac{1molofC_4H_{10}}{58.1222gC_4H_{10}}=0.5506molofC_4H_{10}[/tex]3) Convert the temperature from ºC to K
K = 273.15 K + 45.0ºC = 318.15 K
4) Convert pressure from mmHg to atm
[tex]\text{atm}=728\operatorname{mm}Hg\cdot\frac{1\text{ atm}}{760\text{ mmHg}}=0.9579\text{atm}[/tex]5) Set the equation and solve for V
[tex]\text{PV}=\text{nRT}[/tex][tex]\frac{PV}{P}=\frac{nRT}{P}[/tex][tex]V=\frac{nRT}{P}[/tex]6) Plug in known quantities
n: 0.5506 mol
R: 0.082057 L⋅atm⋅K−1⋅mol−1
T: 318.15 K
P: 0.9579 atm
[tex]V=\frac{(0.5506\text{mol)}(0.082057L\cdot\text{atm}\cdot K^{-1}mol^{-1})(318.15\text{ K)}}{0.9579atm}=\text{ 15.00 L}[/tex]Butane gas will occupy a volume of 15.00 L in the conditions described.
7) Find the volume occupy by butane using R: 8.314 L⋅kPa⋅K−1⋅mol−1
Convert pressure from mmHg to kPa
[tex]\text{kPa}=728\operatorname{mm}Hg\cdot\frac{101325Pa}{760\operatorname{mm}Hg}\cdot\frac{1\text{kPa}}{1000Pa}=97.0587\text{kPa}[/tex]Plug in values
n: 0.5506 mol
R: 8.314 L⋅kPa⋅K−1⋅mol−1
T: 318.15 K
P: 97.0587kPa
[tex]V=\frac{(0.5506mol)(8.314L\cdot kPa\cdot K^{-1}mol^{-1})(318.15\text{ K)}}{97.0587\text{ kPa}}=\text{ 15.01 L}[/tex]
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