Question

Two glass bulbs of equal volume are connected by a narrow tube and filled with a gas at temperature 27°C nsd pressure 77 cm of Hg. One of the bulbs is now kept at temperature 127°C, then new pressure inside the bulb is​

Answer 1

The new pressure inside the bulbs is 88 cm Hg.

Explanation:

Step 1:

The initial temperature of the gas in both the bulbs = 27°C = 27 + 273 = 300 K

The initial pressure in both the bulbs = 77 cm Hg

Both the glass bulbs have equal volumes of gas i.e., total volume = V + V = 2V

Let the total no. of moles be “n”, no. of moles in the first bulb be “n₁” and no. of moles in the second bulb be “n₂”.

∴ The no. of moles in each bulb is given by,

= [Total moles of the gas]/2

= n/2

= [P * 2V] / [2RT] ……. [Ideal Gas Law: PV = nRT]

= PV/ RT

= [77 * V]/[R*300] ……. (i)

Step 2:

It is given that one of the bulbs is kept at a temperature 127°C = 127 + 273 = 400 K

i.e., on heating the second bulb some moles of gas are transferred to the first bulb till the pressure in both the bulbs becomes the same.

Let the new pressure inside the bulbs be denoted as “ P’ ”.

So,

For bulb 1:

n₁ = [P’ * V] / [R * 300]

⇒ n₁ = [P’ *V] / [R*300] ……. (ii)

and,

For bulb 2:

n₂ = [P' * V] / [R * 400]

⇒ n₂ = [P’ * V] / [R * 400] ……. (iii)

Step 3:

Since n = n₁ + n₂

Substituting values of n, n1 & n2 from (i), (ii) & (iii), we get

$\frac{77 * V}{R * 300} * 2 = \frac{P' * V}{R * 300} + \frac{P' * V}{R * 400}$

Cancelling the similar terms

⇒ $\frac{154}{3} = \frac{P'}{3} + \frac{P'}{4}$  

⇒ $\frac{154}{3} = \frac{4P' + 3P'}{12}$

⇒  616 = 7P’

⇒ P’ = $\frac{616}{7}$

⇒ P’ = 88 cm Hg

Thus, the new pressure inside the bulb is 88 cm Hg.

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Answer 2

Explanation:

see the attached file ....