Question

When air at $300\,K$ and $1\,atm$ pressure is compressed to one-twentieth of its original volume and $30\,atm$ pressure, the new temperature will be?

Answer 1

In this question, we will first list out the values given to us and then apply the appropriate values in the gas equation given as $\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$ where the subscripts represent two different conditions. Here $P$ is the pressure, $V$ is the volume and $T$ is the temperature of the gas. The number of moles of gas will remain constant.

Complete step by step answer:
Let the initial pressure be ${P_1}$ , initial volume be ${V_1}$ and the initial temperature be ${T_1}$. In the question it is given that ${P_1} = 1\,atm$ and ${T_1} = 300\,K$. Now the final pressure is given as ${P_2}$ , final volume by ${V_2}$ and the final temperature by ${T_2}$. In the question it is given that ${P_2} = 30\,atm$ and ${V_2} = \dfrac{{{V_1}}}{{20}}$. Now that we have listed the variables and their values as given in the question, we shall move further on writing the appropriate equation in which we can put all the values known to us and get the answer. The gas equation says that for a given number of moles of a particular gas, the quantity $\dfrac{{PV}}{T}$ remains constant. Here P is the pressure, V is the volume and T is the temperature of the gas.

This means that $\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$
Substituting the known values we have,
$\dfrac{{1 \times {V_1}}}{{300}} = \dfrac{{30 \times \dfrac{{{V_1}}}{{20}}}}{{{T_2}}}$
Cancelling out the common terms we have,
$\dfrac{1}{{300}} = \dfrac{{30}}{{20{T_2}}}$
This can be rewritten as
${T_2} = \dfrac{{30}}{{20}} \times 300$
Further solving this we get,
$\therefore {T_2} = 450\,K$

Hence, the new temperature will be ${T_2} = 450\,K$.

Note: The gas equation is based on the assumptions that there are no intermolecular forces between the particles of a gas. Combining the laws of Charles, Boyle and Gay-Lussac gives the combined gas law or the gas equation. Here $T$ is the absolute temperature of the gas and must be always taken in Kelvin. Other quantities can be used in different units they have.