Question

300 litres of ammonia gas at $20^\circ {\rm{C}}$ and 20 atmosphere pressure are allowed to expand in a space of 600 litres capacity and to a pressure of one atmosphere. Calculate the drop in temperature.

Answer 1

We know that, combined gas law expresses the relationship between volume, temperature and pressure of a fixed amount of gas. The expression of combined gas law is,
$\dfrac{{PV}}{T} = k$, where, P is pressure, V is volume, T is temperature and k is constant. Combined gas law is derived from the combination of Charles’ law, Boyle’s law and Gay Lussac’s law.

Complete step by step answer:
Here, ammonia gas of volume 300 litres at 20 atmosphere pressure and $20^\circ {\rm{C}}$ allowed to expand in the space of 600 litres capacity at one atmospheric pressure. So, we have to use the combined gas equation, that is, $\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$ to calculate ${T_2}$. Here, ${P_1}$ is initial pressure, ${V_1}$ is initial volume and ${T_1}$ is initial temperature, ${P_2}$ is final pressure, ${V_2}$ is final volume and ${T_2}$ is final temperature.
Then, we have to calculate the drop in temperature by subtracting ${T_2}$ from ${T_1}$.
Given that, ${P_1}$=20 atm, ${T_1}$=$20^\circ {\rm{C}}$ and ${V_1}$=300 litres
And,
${P_2}$=1 atm , ${V_1}$ = 600 litres
Now, we have to convert ${T_1}$ to Kelvin.
${T_1}$=$20^\circ {\rm{C}} = {\rm{20 + 273}} = {\rm{293}}\,\,{\rm{K}}$
Now, we have to put all the above values in the combined gas equation.
$\Rightarrow \dfrac{{20 \times 300}}{{293}} = \dfrac{{1 \times 600}}{{{T_2}}}$
$\Rightarrow {T_2} = 29.3\,{\rm{K}}$
Now, we have to calculate the drop in temperature.
${T_1} = 293\,{\rm{K}}$ and ${T_2} = 29.3\,{\rm{K}}$
Drop in temperature = ${T_1} - {T_2}$
$\Rightarrow {\rm{Drop}}\,{\rm{in}}\,{\rm{temperature}} = {\rm{293}}\,{\rm{K}} - 29.3\,{\rm{K}} = 263.7\,{\rm{K}}$

Hence, the temperature drop is 263.7 K.

Note: The three gas laws which are combined to give combined gas laws are Boyle's law, Gay-Lussac's law, and Charles' law.
1. According to Boyle's law, at a constant temperature, for a given quantity of gas variation of volume occurs inversely with pressure. The expression of Boyle's law is ${P_1}{V_1} = {P_2}{V_2}$.
2. Charles' law states that, at constant pressure, as volume increases, temperature also increases.
$\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$
3. According to Gay Lussac's law, at constant volume, as pressure increases, temperature increases.
$\dfrac{{{P_1}}}{{{T_1}}} = \dfrac{{{P_2}}}{{{T_2}}}$