Q136 / 160 · TS EAMCET 2008 · Physics

One litre of oxygen at a pressure of $1 \mathrm{~atm}$ and two litres of nitrogen at a pressure of $0.5 \mathrm{~atm}$, are introduced into a vessel of volume $1 \mathrm{~L}$. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is

  • A
    $1.5$
  • B
    $2.5$
  • Correct answer
    C
    $2$
  • D
    $4$

Solution

Ideal gas equation is given by $$ p V=n R T $$ For oxygen, $p=1 \mathrm{~atm}, V=1 \mathrm{~L}, n=n_{\mathrm{O}}$ Therefore Eq. (i) becomes $$ \begin{aligned} & \therefore & 1 \times 1 & =n_{\mathrm{O}_2} R T \\ & \Rightarrow & n_{\mathrm{O}_2} & =\frac{1}{R T} \end{aligned} $$ For nitrogen $p=0.5 \mathrm{~atm}, V=2 \mathrm{~L}, n=n_{\mathrm{N}}$ $$ \begin{array}{rlrl} & \therefore & 0.5 \times 2 & =n_{\mathrm{N}_2} R T \\ \Rightarrow & n_{\mathrm{N}_2} & =\frac{1}{R T} \end{array} $$ For mixture of gas $$ \begin{array}{rlrl} & & p_{\text {mix }} V_{\text {mix }} & =n_{\text {mix }} R T \\ \text { Here, } & n_{\text {mix }} & =n_{\mathrm{O}}+n_{\mathrm{N}_2} \\ & \therefore \quad & \frac{p_{\text {mix }} V_{\text {mix }}}{R T} & =\frac{1}{R T}+\frac{1}{R T} \\ & \Rightarrow & p_{\text {mix }} V_{\text {mix }} & =2 \end{array} $$ Here,

Hence, the correct answer is option: 'C'.

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