Question

If $10g$ of a gas at atmospheric pressure is cooled from ${273^o}C$ to ${0^o}C$, keeping the volume constant, its pressure would become:
(The question has multiple correct options)
(A) $\frac{1}{{273}}atm$
(B) $2atm$
(C) $\frac{1}{2}atm$
(D) $5.05 \times {10^4}N{m^{ - 2}}$

Answer 1

Hint: As per the question, volume is constant that means the question can be solved with the help of the Gay Lussac Law, a thermodynamic law in which the pressure of the system is directly proportional to the temperature when the volume of the system is considered to be constant.

Formula used- $\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}$

Complete Step by Step Solution:
Gay Lussac law: A thermodynamic law in which volume of the system is constant, in such case the pressure of the system varies directly with the temperature of the system and can be mathematically represented as follows:
$P \propto T$
$ \Rightarrow P = kT$
$ \Rightarrow \frac{P}{T} = k$ (Where k is the proportionality constant)

If more than one value of pressure and temperature is given for the system, then the equation can be represented as $\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}\;\;\;...(1)$

According to the question:
${T_1} = {273^o}C = 546K$
${T_2} = {0^o}C = 273K$
${P_1} = 1\;atm$

Substituting the given values in equation (1):
$\frac{1}{{546}} = \frac{{{P_2}}}{{273}}$
$ \Rightarrow {P_2} = \frac{{273}}{{546}} = \frac{1}{2}\;atm$
We know that, $1\;atm = {\rm{101325}}\;N{m^{ - 2}}$

Therefore, ${P_2} = \frac{{{\rm{101325}}}}{2} = 5.05 \times {10^4}N{m^{ - 2}}$
Hence, the pressure after changing the given conditions in the system is $\frac{1}{2}atm = 5.05 \times {10^4}N{m^{ - 2}}$.
Thus, the correct answer is option (C) and (D).

Note: It is important to note that the concept of isochoric process is based on Gay Lussac law according to which no work is done by the system and whatever heat is added to the system is responsible for the increase in the internal temperature of the system whereas the amount of heat rejected by the system will be responsible for the reduction in internal energy of the system.
1. Alcohols and ethers
2. Aldehydes and ketones
3. Carboxylic acids and esters.