Question

If \[1\] litre of gas A is \[600{\text{mm}}\] and \[{\text{0}}{\text{.5L}}\] of a gas B at \[{\text{800mm}}\] are taken in a \[{\text{2L}}\] bulb. The resulting pressure is:
A. \[{\text{1500mm}}\]
B. \[{\text{1000mm}}\]
C. \[{\text{2000mm}}\]
D. \[{\text{500mm}}\]

Answer 1

Hint: First find the no of moles individually for gas A and B and for the resultant mixture using the ideal gas equation. The no of moles will remain the same after mixing both gases, using this condition we can find out the resultant pressure.

Complete step by step answer:
Given,
For gas A, Pressure, \[{P_1} = 600{\text{mm}}\]
And volume, \[{V_1} = 1{\text{L}}\]
For gas B, Pressure, \[{P_2} = 800{\text{mm}}\]
And volume, \[{V_2} = 0.5{\text{L}}\]
Volume of the bulb, \[{V_{final}} = 2{\text{L}}\]

We have the equation for ideal gas as, \[PV = nRT\] (i)
where \[P\] is the pressure, \[V\] is the volume, \[T\] is the temperature, \[n\] is the no of moles of the gas and \[R\] is known as a gas constant.

For A gas, the ideal gas equation will be,
\[{P_1}{V_1} = {n_1}RT \] (ii)
where \[{n_1}\] is the no of moles of gas A.
Now, putting the values of \[{P_1}\] and \[{V_1}\] in equation (ii), we get
\[{P_1}{V_1} = {n_1}RT \\
\Rightarrow (600) \times (1) = {n_1}RT \\
\Rightarrow {n_1} = \dfrac{{600}}{{RT}} \\
\]

For B gas, the ideal gas equation will be,
\[{P_2}{V_2} = {n_2}R{T_2}\] (iii)
where \[{n_2}\] is the no of moles of gas B.
Now, putting the values of \[{P_2}\] and \[{V_2}\] in equation (iii), we get
\[{P_2}{V_2} = {n_2}R{T_2} \\
\Rightarrow (800) \times (0.5) = {n_2}RT \\
\Rightarrow {n_2} = \dfrac{{400}}{{RT}} \\
\]

After the gases are mixed, the total no of moles is
\[n = {n_1} + {n_2}\]
Putting the values \[{n_1}\] and \[{n_2}\], we have
\[n = \dfrac{{600}}{{RT}} + \dfrac{{400}}{{RT}} = \dfrac{{1000}}{{RT}}\] (iv)
Now, the ideal gas equation for the total resultant mixture will be,
\[{P_{final}}{V_{final}} = nRT\]
where \[{P_{final}}\] is the resultant pressure and \[{V_{final}}\] is the volume of the bulb.
\[\therefore n = \dfrac{{{P_{final}}{V_{final}}}}{{RT}}\] (v)
Now, we equate equations (iv) and (v) and we get
\[\dfrac{{{P_{final}}{V_{final}}}}{{RT}} = \dfrac{{1000}}{{RT}}\]
Putting the value of \[{V_{final}} = 2{\text{L}}\], we have
\[\dfrac{{{P_{final}} \times 2}}{{RT}} = \dfrac{{1000}}{{RT}} \\
\Rightarrow {P_{final}} = \dfrac{{1000}}{2} \\
\therefore {P_{final}} = 500\,{\text{mm}} \\
\]

Therefore, the resultant pressure is \[500\,{\text{mm}}\]. Hence, the correct answer is option D.

Note:Here, the units for the quantities of gas A and B are the same. But if the units of quantities of the gases are different then we need to make the units the same before proceeding for calculations as most of us make such mistakes. So, every time before calculations we should check whether the units of the quantities are the same or not.