The Gas Law

Figure (a) above shows a capillary tube with a thread of mercury 2cm long. The length of the air trapped in the tube is 10cm. Find the length of the trapped air if the tube is inverted as shown in figure (b). [Atmospheric pressure = 76cmHg]

Answer:

In case (a), the pressure of the gas trapped in the capillary is equal to the atmospheric pressure + the pressure caused by the mercury thread.

P1 = 76 + 2 = 78cmHg
V1 = 10cm

Incase (b), the pressure of the gas trapped in the capillary is equal to the atmospheric pressure - the pressure caused by the mercury thread.

P2 = 76 - 2 = 74cmHg
V2 = ?

By using Boyle's Lwa's formula

P1V1 = P2V2
(78)(10) = (74)V2
V2 = 10.54cm

The length of the trapped air in figure (b) is 10.54cm

A gas in a container with a constant volume has a pressure of 200,000Pa at a temperature of 30oC. What is the pressure of the gas if the temperature is increased to 60oC?

Answer:

At 30oC,
P1 = 200,000Pa
T1 = 30oC = 273 + 30= 303K
(Note: the temperature must be in Kelvin scale)

At 60oC,
P2 = ?
T2 = 60oC = 273 + 60 = 333K

By using the formula of Pressure Law,

\begin{gathered}

\frac{{P_1 }}

{{T_1 }} = \frac{{P_2 }}

{{T_2 }} \hfill \\

\frac{{(200,000)}}

{{(303)}} = \frac{{P_2 }}

{{(333)}} \hfill \\

P_2  = \frac{{(200,000) \times (333)}}

{{(303)}} = 219802Pa \hfill \\ 

\end{gathered} 

A balloon is filled with 2000cm3 of gas at 27oC. The balloon is immersed in a container filled with water and the water is then heated. If the pressure in the balloon remain constant, find the volume of the gas when its temperature is 57oC.

Answer:

At 27oC,
V1 = 2000cm3
T1 = 27oC = 27 + 273 = 300K

At 57oC,
V2 = ?
T2 = 57oC = 57 + 273 = 330K

By using the formula of Charle's Law,

\begin{gathered}

\frac{{V_1 }}

{{T_1 }} = \frac{{V_2 }}

{{T_2 }} \hfill \\

\frac{{(2,000)}}

{{(30)}} = \frac{{V_2 }}

{{(33)}} \hfill \\

V_2  = \frac{{(2,000) \times (330)}}

{{(300)}} = 2200cm^3  \hfill \\ 

\end{gathered}