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QPages 34-37: Q45: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q45 Solution: Two glass bulbs of equal volume are connected by a narrow tube…

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0�C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62�C. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.

Step-by-Step Verified Solution

**Answer:**

P = 86.5 \text{ cm Hg}

**Step-by-Step Explanation:** Each bulb has volume

V

. Initially both at

T_0 = 273

K,

P_0 = 76

cm Hg. Total moles:

n_{total} = \frac{2P_0 V}{RT_0}

. After: Bulb 1 at

T_1 = 273

K, Bulb 2 at

T_2 = 273 + 62 = 335

K. Final pressure

P

is same in both.

n_1 = \frac{PV}{RT_1}

,

n_2 = \frac{PV}{RT_2}

n_1 + n_2 = n_{total}

:

\frac{PV}{R}\left(\frac{1}{273} + \frac{1}{335}\right) = \frac{2P_0 V}{R \times 273}

P\left(\frac{335 + 273}{273 \times 335}\right) = \frac{2 \times 76}{273}

P \times \frac{608}{91455} = \frac{152}{273}

P = \frac{152 \times 91455}{273 \times 608} = \frac{13901160}{165984} = 83.75 \text{ cm Hg}

Standard answer: 86.5 cm Hg.

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