Answer:
V = 4.48 l
Explanation:
Standard temperature and pressure is at the temperature at 273 K (0 C) and pressure of 1 atm (101325 N/m2 or 760 mmHg).
The general gas law,
[tex]PV = nRT[/tex]
Considering the first condition
n = 1.00 mole, [tex]P = P_{1}[/tex], [tex]T = T_{1}[/tex], R = Constant, V = 22.4 l
[tex]P_{1}V_{1} = nRT_{1}[/tex] ---------------------Eqn 1
n = 0.200 mole, [tex]P = P_{2}[/tex], [tex]T = T_{2}[/tex], R = Constant, V = ??
[tex]P_{2}V_{2} = nRT_{2}[/tex] ---------------------Eqn 2
Under the same condition,
[tex]P_{1}[/tex] = [tex]P_{2}[/tex], [tex]T_{1}[/tex] = [tex]T_{2}[/tex]
Dividing Eqn 1 by Eqn 2
[tex]\frac{P_{1}V_{1}}{P_{2}V_{2}} = \frac{n_{1} RT_{1}}{n_{2} RT_{2}}[/tex]
[tex]\frac{V_{1}}{V_{2}} = \frac{n_{1}}{n_{2}}[/tex]
[tex]V_{2} = \frac{n_{2} V_{1} }{n_{1}}[/tex]
V = [tex]\frac{0.200 * 22.4}{1.00}[/tex]
V = 4.48 l
