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Question

7.1. air is compressed in a cylinder from a pressure of 140 kPa and a temperature of 42 degrees celsuis until the volume is one-quarter of the original volume. The temperature remains constant. calculate the following: 7.2.1. the final pressure of air 7.2.2. the temperature if the air is heated until the pressure in Question 7.2.1. is doubled and the volume remains constant. 7.2. a glass bulb is filled with 0.5m^3 of mucury at a temperature of 25 degrees celsuis. The temperature of the bulb and the temperature of the mercury is increased to 55 degress celsuis. The linear coefficient of expansion of glass in 11X10^-6 degrees celsuis and the volume coefficient of expansion of mercury is 21X10^-16 degrees celsuis. Calculate the overflow of mercury.

Answer 1

(7.2.1)

The Pressure of the gas = 140 kPa for one-fourth of the volume.

Let v be the original volume then one-fourth of the volume = v/4

From Boyle's Law of gases, when the temperature is constant then

140 * (V/4) = PV

P = 140/4

P = 35 kPa.

Final Pressure = 35 kPa.

Answer 2

(7.2.2)

Temperature = 42°C

T = 42 + 273 = 315 K

Let P be the pressure.

Temperature when Pressure is doubled = T.

From Gay - Lussac's Law of gases, when the temperature is constant then

T = 2 * 315

T = 630 K

T = 630 - 273 = 357

T = 357°C

Temperature at which the pressure is doubled = 357°C.

Answer 3

(7.2)

The volume of the mercury = 0.5 m³

Initial Temperature = 25°C

Final Temperature = 55°C

Volume Co-effiecient of Expansion of mercury = 21 * 10^-16

Therefore the overflow of mercury is 3.15*10^-14 m³.