Solveeit Logo

Question

Question: A cylinder of 5 L capacity, filled with air at NTP is connected with another evacuated cylinder of 3...

A cylinder of 5 L capacity, filled with air at NTP is connected with another evacuated cylinder of 30 L capacity. The resultant air pressure in both the cylinder will be:
A) 10.810.8 cm of Hg
B) 14.914.9 cm of Hg
C) 21.821.8 cm of Hg
D) 38.838.8 cm of Hg

Explanation

Solution

We know that the product of pressure and volume of a fixed amount of a gas remains constant at a constant temperature. So, the pressure and volume of the cylinder will be equal to resultant pressure and volume. Use the formula to find the value of resultant pressure.
Formula used:
PV = constant{\text{PV = constant}}
Here, P{\text{P}}= pressure of the gas
V{\text{V}}= volume of the gas

Complete step by step answer:
Let us consider the cylinder with 5 L capacity as cylinder 1.
We can write the formula for cylinder 1 as:
  P1V1 = constant\;{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = constant}} ……(equation 1)
Here, P1{{\text{P}}_{\text{1}}} = Pressure at NTP
= 76 cm of Hg
V1{{\text{V}}_{\text{1}}} = Volume of cylinder
= 5 L
Now, let us consider the resultant cylinder (i.e. cylinder of 5 L capacity connected with the cylinder of 30 L capacity) as cylinder 2.
So, we can write the formula for cylinder 2 as:
  P2V2 = constant\;{{\text{P}}_2}{{\text{V}}_2}{\text{ = constant}} …… (equation 2)
Here, P2{{\text{P}}_{\text{2}}} = Pressure of resultant air in both the cylinder
V2{{\text{V}}_{\text{2}}} = Volume of resultant cylinder
Hence,
V2=5+30 =35 L  {{\text{V}}_{\text{2}}} = 5 + 30 \\\ = 35{\text{ L}} \\\
Comparing equation 1 and equation 2 we get:
P1V1=P2V2{P_1}{V_1} = {P_2}{V_2}
Let us substitute the respective value we get:
76 (cm of Hg)×5(L)=P2×35(L){\text{76 (cm of Hg)} \times 5 (L) = }{{\text{P}}_{\text{2}}}{\text{} \times 35 (L)}
By rearranging the equation we get:
76 (cm of Hg)×5(L)35 (L) = P2\dfrac{{{\text{76 (cm of Hg)} \times {5 (L)}}}}{{{\text{35 (L)}}}}{\text{ = }}{{\text{P}}_{\text{2}}}
As volume terms are present in numerator and denominator, we should cancel the term to get:
76 (cm of Hg)×1(L)7(L) = P2\dfrac{{{\text{76 (cm of Hg)} \times 1 ({L})}}}{{{\text{7} ({L})}}}{\text{ = }}{{\text{P}}_{\text{2}}}
Now, by dividing the pressure value by 7 we get:
P2 = 10.8 cm of Hg{{\text{P}}_{\text{2}}}{\text{ = 10}}{\text{.8 cm of Hg}}

Therefore, we can conclude that the correct answer to this question is option A.

Note:
We can get confused between the selection of P2{{\text{P}}_{\text{2}}} as the pressure of a cylinder having 30 L capacity or the resultant pressure of cylinder 1 plus cylinder 2. As cylinder 2 having 30 L is evacuated, the pressure inside it is zero and thus, we choose P2{{\text{P}}_{\text{2}}} as resultant pressure.