Question

The two bulbs of volume 5 litre and 10 litre containing an ideal gas at 9 atm and 6 atm respectively are connected what is the final pressure in the two bulbs if the temperature remains constant

Answer 1

P1V1 + P2V2 = P(V1 + V2)
5*9 + 10*6 = P(15)
45+60= 15P
105÷15 = P
P = 7 atm

Answer 2

Pressure of the combined gases is 7 atm

Solution:

Let us assume the gases in two bulbs are X and Y.

We are given that the quantity of X = 5l L

Atmospheric pressure of X = 9 atm

Quantity of y = 10 L

Atmospheric pressure of Y = 6 atm

We also know that the bulbs are connected and the temperature is constant.

We can calculate the total volume of the mixed gas by using PV (Pressure × Volume)

So, the total volume of mixture of gas = ((5×9) + (10×6))

= 105 L

Total quantity of gases= (5+10)l = 15 L

$Pressure \ of \ the \ combined \ gases = \frac{Volume \ of \ combines \ gases}{Quantity \ of \ combined \ gases}$

(This is according to the Boyle’s Law which defines that the volume of a confined gas is indirectly proportional to its volume).

$=\frac{105}{15}$

Pressure of the combined gases = 7 atm