
A gaseous mixture containing 0.35 g of N2 and 5600 ml of O2 at STP is kept in 5 litres of flask at 300K. The total pressure of the gaseous mixture is:
A. 1.293 atm
B. 1.2315 atm
C. 12.315 atm
D. 0.616 atm
Answer
565.8k+ views
Hint: In a mixture of gases, each constituent gas has a partial pressure which is the notional pressure of that constituent gas if it alone occupied the entire volume of the original mixture at the same temperature, i.e. If a mixture contain 3 gases such as A, B, and C then the total pressure of that mixture will be the sum of their partial pressure of each gases
\[\Rightarrow {{P}_{Total}}={{p}_{A}}+{{p}_{B}}+{{p}_{C}}\]
Complete step by step answer:
In the question we have given that 0.35 g N2 and 5600 ml O2, hence the total pressure of the container will be,
\[\Rightarrow {{P}_{Total}}={{p}_{{{N}_{2}}}}+{{p}_{{{O}_{2}}}}\]
\[{{p}_{{{N}_{2}}}}\]= Partial pressure of N2
\[{{p}_{{{O}_{2}}}}\]= Partial pressure of O2
Here we have partial pressure of nitrogen is equal to,
\[\Rightarrow {{p}_{{{N}_{2}}}}=\dfrac{{{n}_{{{N}_{2}}}}\times R\times T}{V}\]
\[\Rightarrow {{n}_{{{N}_{2}}}}=\]No of moles of N2 = 0.35÷28 (molecular mass of nitrogen is 28 g)= 0.0125 mole
\[\Rightarrow R=\]Ideal gas constant= 0.0821 atm L/mol K
\[\Rightarrow T=\]Temperature= 300 K
\[\Rightarrow V=\]Total volume= 5 Litre
Substitute these values in the equation, then we get
\[\Rightarrow {{p}_{{{N}_{2}}}}=\dfrac{0.0125\times 0.0821\times 300}{5}=0.061atm\]
Like this we have to find the partial pressure of 5600 ml of O2 at STP. 1 mole oxygen contains 22400 ml of oxygen, here we have 5600 ml. So 5600 ml of oxygen contain,
\[\Rightarrow \dfrac{5600}{22400}mole=0.25mole\]
Hence we get the no of mole of oxygen as 0.25 moles. Then we have partial pressure of oxygen is equal to,
\[\Rightarrow {{p}_{{{O}_{2}}}}=\dfrac{{{n}_{{{O}_{2}}}}\times R\times T}{V}\]
\[\Rightarrow {{n}_{{{O}_{2}}}}\]= 0.25 mole
Substitute these values in the equations. Then we get,
\[\Rightarrow {{p}_{{{O}_{2}}}}=\dfrac{0.25\times 0.0821\times 300}{5}=1.231atm\]
Hence the total pressure of the mixture is equal to
\[\Rightarrow {{P}_{Total}}={{p}_{{{N}_{2}}}}+{{p}_{{{O}_{2}}}}\]
\[\Rightarrow {{P}_{total}}=0.061+1.231=1.293atm\]
The total pressure is 1.293 atm .Hence, the correct answer is option A.
Note:
Please note that the mixture should contain non-reactive gas. They should not react with each other. If once they react the individual gases give rise to a new product. For a mixture of nitrogen and hydrogen the partial pressure concept can not be applied. Because they will react to each other and form ammonia gas.
\[\Rightarrow {{P}_{Total}}={{p}_{A}}+{{p}_{B}}+{{p}_{C}}\]
Complete step by step answer:
In the question we have given that 0.35 g N2 and 5600 ml O2, hence the total pressure of the container will be,
\[\Rightarrow {{P}_{Total}}={{p}_{{{N}_{2}}}}+{{p}_{{{O}_{2}}}}\]
\[{{p}_{{{N}_{2}}}}\]= Partial pressure of N2
\[{{p}_{{{O}_{2}}}}\]= Partial pressure of O2
Here we have partial pressure of nitrogen is equal to,
\[\Rightarrow {{p}_{{{N}_{2}}}}=\dfrac{{{n}_{{{N}_{2}}}}\times R\times T}{V}\]
\[\Rightarrow {{n}_{{{N}_{2}}}}=\]No of moles of N2 = 0.35÷28 (molecular mass of nitrogen is 28 g)= 0.0125 mole
\[\Rightarrow R=\]Ideal gas constant= 0.0821 atm L/mol K
\[\Rightarrow T=\]Temperature= 300 K
\[\Rightarrow V=\]Total volume= 5 Litre
Substitute these values in the equation, then we get
\[\Rightarrow {{p}_{{{N}_{2}}}}=\dfrac{0.0125\times 0.0821\times 300}{5}=0.061atm\]
Like this we have to find the partial pressure of 5600 ml of O2 at STP. 1 mole oxygen contains 22400 ml of oxygen, here we have 5600 ml. So 5600 ml of oxygen contain,
\[\Rightarrow \dfrac{5600}{22400}mole=0.25mole\]
Hence we get the no of mole of oxygen as 0.25 moles. Then we have partial pressure of oxygen is equal to,
\[\Rightarrow {{p}_{{{O}_{2}}}}=\dfrac{{{n}_{{{O}_{2}}}}\times R\times T}{V}\]
\[\Rightarrow {{n}_{{{O}_{2}}}}\]= 0.25 mole
Substitute these values in the equations. Then we get,
\[\Rightarrow {{p}_{{{O}_{2}}}}=\dfrac{0.25\times 0.0821\times 300}{5}=1.231atm\]
Hence the total pressure of the mixture is equal to
\[\Rightarrow {{P}_{Total}}={{p}_{{{N}_{2}}}}+{{p}_{{{O}_{2}}}}\]
\[\Rightarrow {{P}_{total}}=0.061+1.231=1.293atm\]
The total pressure is 1.293 atm .Hence, the correct answer is option A.
Note:
Please note that the mixture should contain non-reactive gas. They should not react with each other. If once they react the individual gases give rise to a new product. For a mixture of nitrogen and hydrogen the partial pressure concept can not be applied. Because they will react to each other and form ammonia gas.
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