Question

A vessel containing 10 liters of air under a pressure of 1MPa is connected to a 4-liter empty vessel. The final air pressure in the vessel assumes that the process is isothermal.
A) (7/5) MPa
B) (5/7) MPa
C) 1MPa
D) 10MPa

Answer 1

Hint: In isothermal process PV both changes keeping the T constant. If the system is closed then no. of moles would also be conserved.

Formula used:
In the isothermal process, for a closed system we can say: ${P_1}{V_1} = {P_2}{V_2}$=constant …… (1)
Where P represents pressure and V represents a corresponding volume.
Ideal gas law, $PV = nRT$
Where $P$ is pressure, $V$ is volume, $R$ is Rydberg Constant, $T$ is temperature and $n$ is the number of moles of an ideal gas in the system.

Complete step by step answer:
Given,
Initial Pressure in the vessel (closed): ${P_1} = 1MPa$
Initial Volume occupied by the ideal gas in given system: ${V_1} = 10litre$

Step 1:
For the isothermal process, PV= constant
So, using equation (2), we can say initially PV is constant for given T which correspond to no. of moles in the system
${P_1}{V_1} = (1MPa)(10litre)$ …… (3)

Step 2:
${V_2}$ be the final volume after the isothermal process= Volume of the entire container (old + newly added)
$ \Rightarrow {V_2} = (10 + 4)$litres …… (4)

Step 3:
Let Final pressure be ${P_2}$ and Volume corresponding to this pressure is ${V_2}$
Hence, by using equation (1) and value from equation (4) we can say-
\[ {P_1}{V_1} = {P_2}{V_2} \]
$\Rightarrow (1MPa)(10litre) = {P_2}(10 + 4)litre $
$\Rightarrow {P_2} = \dfrac{{(1MPa)(10litre)}}{{(10 + 4)litre}}$
$\Rightarrow {P_2} = \dfrac{5}{7}MPa $

The final air pressure in the vessel $\dfrac{5}{7}MPa$. So, option (B) is correct.

Note:
Here we have assumed the vessel to be closed. This is so, because, if the vessel is open then there would not be any pressure due to gas. It would flee away, Hence, no pressure would have been there as ideal gas is not interacting with each other so due to elastic collision it would fly off the vessel.