Question

The stopcock, connecting the two bulbs of volumes 5 litres and 10 litres containing an ideal gas at 9 atm and 6 atm respectively, is opened. What is the final pressure in the two bulbs if the temperature remained the same ?

Answer 1

$\large\underline\mathrm\green{Given:-}$

• P1 = 9 atm.
• V1 = 5 litres.
• P2 = 6 atm.
• V2 = 10 litres.

$\large\underline\mathrm\red{Solution}$

• For the first bulb,

p1V1 = n1RT

• For the second bulb,

p2V2 = n2RT

$\large\underline\mathrm\red{On \: adding:}$

p1V1 + p2V2 = (n1 + n2)RT.

When the stopcock is opened, the total volume is (V1 + V2) and total no. of moles is (n1 + n2). Suppose the equilibrium pressure at each bulb is p atm.

$\large\underline\mathrm\red{Then,}$

p(V1 + V2) = (n1 + n2)RT

$\large\underline\mathrm\red{From \: equations \: (1) \: and \: (2), \: we \: get,}$

p1V1 + p2V2 = p(V1 + V2)

$\large\underline\mathrm\red{Substituting:}$

$\implies$9 × 5 + 6 × 10 = p × 15

$\implies$p = 7 atm.

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