1. Tardigrade
  2. Question
  3. Chemistry
  4. The internal energy (U) of an ideal gas is plotted against volume for a cyclic process A B C D A, as shown in the figure. <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/chemistry/9665d284427fee84a1c690ff89b05cad-.png /> The temperature of the gas at B and C are 500 K and 300 K, respectively, The heat absorbed by the gas (in cal/ mol) in this cyclic process, is:

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Q. The internal energy ($U$) of an ideal gas is plotted against volume for a cyclic process $A B C D A$, as shown in the figure.
image
The temperature of the gas at $B$ and $C$ are $500 \,K$ and $300 \,K$, respectively, The heat absorbed by the gas (in cal/ mol) in this cyclic process, is:

Thermodynamics

Solution:

$q = q _{ AB }+ q _{ BC }+ q _{ CD }+ q _{ DA }$
$= nRT _{ B } \cdot \ln \frac{2 V _{0}}{ V _{0}}+ nC _{ v , m }\left( T _{ C }- T _{ B }\right)+ nRT _{ c } \ln \frac{ V _{0}}{2 V _{0}}$
$+ nC _{ v , m }\left( T _{ A }- T _{ D }\right)$
$= nR \ln 2\left( T _{ B }- T _{ C }\right)$
$=1 \times 2 \times 0.7 \times 200=280\, cal / mol$