Question

A sample of $ {{\text{O}}_2} $ gas and a sample of hydrogen gas both have the same no. of moles, the same volume and same pressure. Assuming them to be perfect gases, the ratio of the temperature of oxygen gas to the temperature of hydrogen gas is:
(A) $ 1:1 $
(B) $ 1:2 $
(C) $ 2:1 $
(D) $ 1:4 $

Answer 1

Hint All ideal (perfect) gases, irrespective of the particular element, behave identically. Here we have to compare the temperature values predicted by the ideal gas relation to get the answer.
Formula used: In this solution we will be using the following formula;
 $ \Rightarrow PV = nRT $ where $ P $ is the pressure of the gas, $ V $ is its temperature, $ n $ is the number of moles, $ R $ is the universal gas constant, and $ T $ is the temperature of the gas.

Complete step by step answer
Earlier experiments, such as Boyle’s experiment, Charles’s experiment, identified that gases irrespective of element behave identically under the same condition. These experiments led to a famous law, well-known today as the ideal gas law. This law can be said to state that the pressure of a given mass of gas is directly proportional to the absolute temperature and inversely proportional to the volume of the gas.
It is mathematically given as:
 $ \Rightarrow P \propto \dfrac{T}{V} $
 $ \Rightarrow P = k\dfrac{T}{V} $
Where $ P $ is the pressure of the gas, $ V $ is its temperature, $ T $ is the temperature of the gas, and $ k $ is a constant of proportionality. In some cases, it is given as $ nR $ where $ n $ is the number of moles, and $ R $ is the universal gas constant.
Hence, we have that
 $ \Rightarrow PV = nRT $
Now, in the question, both the oxygen and hydrogen are to be assumed to be perfect gases, hence for oxygen, we have
 $ \Rightarrow {P_o}{V_o} = {n_o}R{T_o} $ where the subscript $ o $ shows the variables are for oxygen.
Similarly, for hydrogen gas,
 $ \Rightarrow {P_h}{V_h} = {n_h}R{T_h} $ .
We shall compare the two by dividing, hence
 $ \Rightarrow \dfrac{{{P_o}{V_o}}}{{{P_h}{V_h}}} = \dfrac{{{n_o}R{T_o}}}{{{n_h}R{T_h}}} $
 $ \Rightarrow \dfrac{{{P_o}{V_o}}}{{{P_h}{V_h}}} = \dfrac{{{n_o}{T_o}}}{{{n_h}{T_h}}} $
Now, according to the question, the pressure, volume and number of moles of both substances are equal, i.e. $ {P_o} = {P_h} = P $ , $ {V_o} = {V_h} = V $ , $ {n_o} = {n_h} = n $ . Hence,
 $ \Rightarrow \dfrac{{PV}}{{PV}} = \dfrac{{n{T_o}}}{{n{T_h}}} $
 $ \Rightarrow \dfrac{{{T_o}}}{{{T_h}}} = \dfrac{1}{1} $
Hence, the ratio is $ 1:1 $
Thus, the correct answer is A.

Note
In reality, although Boyle and Charles concluded at this law and the question has us assume it, the ideal gas is fatally wrong. There is no such thing as a perfectly ideal gas. However, at low temperatures and pressures (like the ones used by Boyle and Charles) the ideal gas law is an exceptional approximation to the real condition.