Question

A cylindrical glass beaker of height 1.272 m rests on a table. The bottom half of the beaker is filled with a gas, and the top half is filled with liquid mercury that is exposed to the atmosphere. The gas and mercury do not mix because they are separated by a frictionless, movable piston of negligible mass and thickness. The initial temperature is 276 K. The temperature is increased until a value is reached when one-half of the mercury has spilled out. Ignore the thermal expansion of the glass and mercury, and find this temperature in kelvins.

Answer 1

Given that :

initial temperature, T₁ = 276 K

height of cylindrical beaker, h = 1.272 m

Ignore the thermal expansion of the glass and mercury.

using a combined gas laws,

P₁ V₁ / T₁ = P₂ V₂ / T₂                                      { eq.1 }

where, V = A d

then, P₁ A d₁ / T₁ = P₂ A d₂ / T₂       

rearranging an above eq.

T₂ = ( P₂ / P₁) (d₂ / d₁) T₁                                   { eq.2 }

the initial pressure is given as :

P₁ = P + gh₁                                                      { eq.3 }

where, P = constant pressure = 1.01 x 10⁵ Pa

= density of mercury = 1.36 x 10⁴ kg/m³

g = acceleration due to gravity = 9.8 m/s²

h₁ = height of beaker = h / 2 = (1.272 m) / 2 = 0.636 m

inserting these values in eq.3,

P₁ = (1.01 x 10⁵ Pa) + (1.36 x 10⁴ kg/m³) (9.8 m/s²) (0.636 m)

P₁ = (1.01 x 10⁵ Pa) + (8.47 x 10⁴ Pa)

P₁ = 185700 Pa

and final pressure is given as :

P₂ = P + gh₂                                                   { eq.4 }

where, h₂ = h₁ / 2 = (0.636 m) / 2 = 0.318 m

P = constant pressure = 1.01 x 10⁵ Pa

inserting the values in eq.4,

P₂ = (1.01 x 10⁵ Pa) + (1.36 x 10⁴ kg/m³) (9.8 m/s²) (0.318 m)

P₂ = (1.01 x 10⁵ Pa) + (4.23 x 10⁴ Pa)

P₂ = 143300 Pa

Now, inserting all these values in eq.2,

T₂ = ( P₂ / P₁) (d₂ / d₁) T₁        

T₂ = [(143300 Pa) / (185700 Pa)] (d₂ / d₁) (276 K)

where, d₁ = 0.636 m   and d₂ = (0.636 m) + (0.318 m) = 0.954 m

then, T₂ = [(143300 Pa) / (185700 Pa)] [(0.954 m) / (0.636 m)] (276 K)

T₂ = (0.7716) (1.5) (276 K)

T₂ = 319.4 K