Question

A certain quantity of a gas occupies a volume of 0.1 litre when collected over water at ${27^0}C$ and pressure 1.67 atm. The same amount of gas occupied 0.14 litre at 1 atm, ${7^0}C$ in dry condition. Calculate the aqueous tension (in atm) at ${27^0}C$.

Answer 1

Hint: The combined gas law is an amalgamation of the three previously known laws which are- Boyle’s law where \[PV{\text{ }} = {\text{ }}K\], Charles law where \[\dfrac{V}{T} = K\], and Gay-Lussac’s law where \[\dfrac{P}{T} = K\]. Therefore, the formula of combined gas law is \[\dfrac{{PV}}{T} = K\]. It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.
Where,
P = pressure,
T = temperature,
V = volume,
K is constant.

Complete step by step solution:
We know that,
\[{P_{drygas}} = {P_{observed}} - \] aqueous tension
Aqueous tension \[ = \;{P_{observed}} - {P_{drygas}}\]
Now, \[{P_{observed}} = 1.67\;atm\]
\[{P_{drygas}}_\; = ?\]
According to combined gas law
P1V1T1 = P2V2T2
Given in the question are:
\[{P_1}\; = ?\]
\[{P_2}\; = 1atm\]
${T_1} = 280K$
${T_2} = 300K$
${V_1} = 0.1l$
${V_2} = 0.14l$
Therefore, Pressure of dry gas,
${P_1} = \dfrac{{{P_2}{V_2}}}{{{T_2}}} \times \dfrac{{{T_1}}}{{{V_1}}}$
$ \Rightarrow {P_1} = \dfrac{{1atm \times 0.14l \times 280K}}{{300K \times 0.1}}$
$ \Rightarrow {P_1} = 1.31atm$
Therefore, Aqueous tension
\[ = {P_{observed}} - {P_{drygas}}\]
\[ = 1.67 - 1.31\]
\[ = 0.36\;atm\]

Thus, the aqueous tension is 0.36 atm.

Note: The combined gas law has practical uses when dealing with gases at ordinary temperatures and pressures. Like other gas laws based on ideal behaviour, it becomes less accurate at high temperatures and pressures. The law is used in thermodynamics and fluid mechanics. For example, it can be used to calculate pressure, volume, or temperature for the gas in clouds to forecast weather.