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Question: If \(500\;ml\) of gas A at \(1000\;torr\), and \(1000\;ml\) of gas B at \(800\;torr\) are placed in ...

If 500  ml500\;ml of gas A at 1000  torr1000\;torr, and 1000  ml1000\;ml of gas B at 800  torr800\;torr are placed in a 2L2L container, the final pressure will be
A. 100torr100\,torr
B. 650torr650\,torr
C. 1800torr1800\,torr
D. 2400torr2400\,torr

Explanation

Solution

We can use Dalton’s law of partial pressure here to solve the problem. It states that when a mixture of two or more non-reacting gases are enclosed in a container then the total pressure exerted by the gaseous mixture is equal to the sum of partial pressure of the individual gases.

Complete step by step solution:
From the hint we understood what Dalton's law of partial pressure was.
So, the mathematical expression for this is,
Ptotal=P1+P2{{\text{P}}_{{\text{total}}}} = {{\text{P}}_1} + {{\text{P}}_2}
Where Ptotal{{\text{P}}_{{\text{total}}}} is total pressure exerted
P1{{\text{P}}_1} and P2{{\text{P}}_2} are the partial pressure of gases
We know that from the Ideal gas equation,
PV=nRTPV = nRT
Where PP is the pressure
VV is the volume
nn is the number of moles
RR is universal gas constant
TT is the temperature
Here, we are assuming that the temperature remains constant.
Then, we can write it as
Ptotal=ntotalRTVtotal{P_{total}} = \dfrac{{{n_{total}}RT}}{{{V_{total}}}}
Where ntotal{n_{total}} is the total number of moles of gases.
Since the total number of moles in the container will be equal to the sum of individual number of moles we can write
ntotal=n1+n2{n_{total}} = {n_1} + {n_2}
Where n1{n_1} and n2{n_2} are the number of moles of individual gases.
And Vtotal=2L{V_{total}} = 2L is given
Now we have to find n1{n_1} and n2{n_2}
Applying the ideal gas law here we get
n1=P1V1RT\Rightarrow {n_1} = \dfrac{{{P_1}{V_1}}}{{RT}} and n2=P2V2RT{n_2} = \dfrac{{{P_2}{V_2}}}{{RT}}
Substituting the values in Ptotal{P_{total}}
Ptotal=(P1V1RT+P2V2RT)RTVtotal=P1V1+P2V2Vtotal\Rightarrow {P_{total}} = \dfrac{{\left( {\dfrac{{{{\text{P}}_1}{{\text{V}}_1}}}{{{\text{RT}}}} + \dfrac{{{P_2}{V_2}}}{{RT}}} \right)RT}}{{{V_{total}}}} = \dfrac{{{P_1}{V_1} + {P_2}{V_2}}}{{{V_{total}}}}
It is given that,
P1=500ml  and  V1=1000torr{P_1} = 500\,ml\,\;and\;\,{V_1} = 1000\,torr
And P2=1000ml  and  V2=800torr{P_2} = 1000\,ml\;and\;{V_2} = 800\,torr
Vtotal=2L=2000ml{V_{total}} = 2L = 2000\,ml
Substituting these values we get,
Ptotal=500×1000+1000×80002000\Rightarrow {P_{total}} = \dfrac{{500 \times 1000 + 1000 \times 8000}}{{2000}}
Ptotal=650torr\Rightarrow {P_{total}} = 650\,torr

**Therefore the final pressure will be 650torr650 \,torr i.e., option (b) is correct

Note:**
If the temperature is constant, we can modify the ideal equation to
P1V1n1=P2V2n2\dfrac{{{P_1}{V_1}}}{{{n_1}}} = \dfrac{{{P_2}{V_2}}}{{{n_2}}}
Where PP, VV and nn are all the same as mentioned above.