Question

$3.2g$ of oxygen (at wt=16) and $0.2$ of hydrogen $(atwt=1$ are placed in a $1.12L$ flask at $0^{o}C$. FInd the total pressure of the gas mixture.

Answer 1

Solution:-

From ideal gas equation-

$PV=nRT.....\left(1\right)$

whereas,

$P=$ pressure of the gas

$V=$ volume it occupies

$n=$ number of moles of gas present in the sample $=\frac{\text{mass of gas}\left(m\right)}{\text{molar mass of gas}\left(M\right)}$

$R=$ universal gas constant $=0.0821atmL{mol}^{-1}{K}^{-1}$

$T=$ absolute temperature of the gas

Therefore,

For $H_2$	For $O_2$
Given mass of $H_2=0.2g$ Molar mass of $H_2=1g$ $n=\frac{0.2}{1}=0.2\text{mole}$ $P=p_{H_2}=?$ $V=1.12L$ $T=0℃=273K$ From ${eq}^n\left(1\right)$, we have $p_{H_2}\times 1.12=0.2\times 0.0821\times 273$ $\Rightarrow p_{H_2}\approx 4\text{atm}$	Given mass of $O_2=3.2g$ $n=\frac{3.2}{16}=0.2\text{mole}$ Molar mass of $O_2=16g$ $P=p_{O_2}=?$ $V=1.12L$ $T=0℃=273K$ From ${eq}^n\left(1\right)$, we have $p_{O_2}\times 1.12=0.2\times 0.0821\times 273$ $\Rightarrow p_{O_2}\approx 4\text{atm}$

According to Dalton's law of partial pressure, total pressure of the mixture is equal to the sum of partial pressure of the individual gases.

$\therefore p_{mix.}=p_{H_2}+p_{O_2}$

$\Rightarrow p_{mix.}=(4+4)\text{atm}=8\text{atm}$

Hence, the final pressure of the mixture in will be $8\text{atm}$.