
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
163.5k+ views
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.
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Instantaneous Velocity - Formula based Examples for JEE

JEE Advanced 2025 Notes
