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

Question

One litre of oxygen at a pressure of 1atm and two litres of nitrogen at a pressure of 0.5atm are introduced into a vessel of volume 1L . If there is no change in temperature, the final pressure of the mixture of gasses (in atm ) is (A) 1.5 (B) 2.5 (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

Ideal gas equation is given by, pV=nRT ...(i) For oxygen, p=1atm,V=1L,n=nO2 Therefore Eq. (i) becomes 1× 1=nO2RT ⇒ nO2=(1/R T) For nitrogen p=0.5atm,V=2L,n=nN2 ∴ 0.5× 2=nN2RT ⇒ nN2=(1/R T) For mixture of gasses, Pm i x Vm i x=nm i xRT Here nm i x=nO2+nN2 ∴ (Pm i x Vm i x/R T)=(1/R T)+(1/R T) ⇒ Pm i xVm i x=2 Pmix=2

Q. One litre of oxygen at a pressure of $1 a t m$ and two litres of nitrogen at a pressure of $0.5 a t m$ are introduced into a vessel of volume $1 L$ . If there is no change in temperature, the final pressure of the mixture of gasses (in $a t m$ ) is

$1.5$

$2.5$

$2$

$4$

Q. One litre of oxygen at a pressure of 1atm and two litres of nitrogen at a pressure of 0.5atm are introduced into a vessel of volume 1L . If there is no change in temperature, the final pressure of the mixture of gasses (in atm ) is

1.5

2.5

2

4

Q. One litre of oxygen at a pressure of $1 a t m$ and two litres of nitrogen at a pressure of $0.5 a t m$ are introduced into a vessel of volume $1 L$ . If there is no change in temperature, the final pressure of the mixture of gasses (in $a t m$ ) is

$1.5$

$2.5$

$2$

$4$