Answer to Question #123197 in General Chemistry for Dyana

Question

A closed gas cylinder contains 0.500 mole H2, 0.300 mole O2, 1.200 mole CO2 at 25 °C and pressures 2.00 atm. Calculate the volume of the cylinder. Then, calculate the partial pressure of O2 inside the cylinder in question.

((R=0.082 L atm mol^-1 K^-1)

Answer 1

Solution.

$n_{H_2}=0.500mol;$

$n_{O_2}=0.300mol;$

$n_{CO_2}=1.200mol;$

$T=25^oC=298K;$

$P_{total}=2 atm;$

$P_{total}V=n_{total}RT;$

$V=\dfrac{n_{total}RT}{P_{total}};$

$n_{total}=n_{H_2}+n_{O_2}+n_{CO_2};$

$V=\dfrac{(0.500+0.300+1.200)mol\sdot0.082L \sdot atm \sdot mol^{-1}K^{-1}\sdot298K}{2atm}=24.44L;$

$P_{O_2}=n_{O_2}\dfrac{RT}{V};$

$P_{O_2}=0.300mol \dfrac{0.082L\sdot atm\sdot mol^{-1}\sdot K^{-1}\sdot298K}{24.44L}=0.3atm;$

Answer: $V=24.44L;$

$P_{O_2}=0.3atm;$