One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are added into a vessel of volume 1 L . If the change in temperature is zero, then find the final pressure of the mixture of gas (in atm ) will be:

Question

One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are added into a vessel of volume 1 L . If the change in temperature is zero, then find the final pressure of the mixture of gas (in atm ) will be: (A) 1.5  (B) 2.5  (C) 2  (D) 4

Tardigrade Admin · Accepted Answer

The Ideal gas equation is given by PV=nRT ... (i) For oxygen, P=1atm , V=1L , n=n texto2 Therefore, Eq. (i) becomes ∴ 1× 1=nO2RT ∴ nO2=(1/R T) For nitrogen P=0.5atm , V=2L , n=nN2 Therefore 0.5× 2=nN2RT ⇒ nN2=(1/R T) For the mixture of gas PmixVmix=nmixRT â Here, n textmix=n texto2+n textN2 (P textmix V textmix/R T)=(1/R T)+(1/R T) ⇒ PmixVmix=2 ⇒ Pmix=2 atm

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 added into a vessel of volume $1 L$ . If the change in temperature is zero, then find the final pressure of the mixture of gas (in $a t m$ ) will be:

$1.5$

$2.5$

$2$

$4$

Q. One litre of oxygen at a pressure of 1atm and two litres of nitrogen at a pressure of 0.5atm are added into a vessel of volume 1L . If the change in temperature is zero, then find the final pressure of the mixture of gas (in atm ) will be:

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 added into a vessel of volume $1 L$ . If the change in temperature is zero, then find the final pressure of the mixture of gas (in $a t m$ ) will be:

$1.5$

$2.5$

$2$

$4$