Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Calculate the change in entropy (S) when 2 moles of an ideal gas expand isothermally and reversibly from a volume of 10 L to a volume of 20 L at a temperature of 300 K.<

Calculate the change in entropy (S) when 2 moles of an ideal gas expand isothermally and reversibly from a volume of 10 L to a volume of 20 L at a temperature of 300 K.

Option: 1

 -12 JK
 


Option: 2

-11.52 J-K
 


Option: 3

 -11.52 J/K
 


Option: 4

 -11.52 kJ/K


Answers (1)

best_answer

The change in entropy (?S) for an isothermal and reversible expansion of an ideal gas can be calculated using the equation:

\mathrm{\Delta S=n R \ln \left(\frac{V_f}{V_i}\right)}

where n is the number of moles of the gas, R is the ideal gas constant (8.314 J/(mol·K)), \mathrm{V_f} is the final volume, and \mathrm{V_i} is the initial volume.

Given:

\mathrm{\begin{aligned} n & =2 \text { moles } \\ R & =8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ V_f & =20 \mathrm{~L} \\ V_i & =10 \mathrm{~L} \end{aligned}}

Substituting the values and calculating:

\mathrm{\Delta S=2 \times 8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \times \ln \left(\frac{20 \mathrm{~L}}{10 \mathrm{~L}}\right)}

Therefore, the change in entropy (?S) is approximately ?S = −11.52 J/K.
So, option c is correct

 

 

 

Posted by

Ritika Harsh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question