Question 996: Thermodynamics & KTG

Question

A cyclic process for 1 mole of an ideal is shown in the $V-T$ diagram. The work done in $A B, B C$ and $C A$ respectively is
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Accepted Answer

In the cyclic process for 1 mole of an ideal gas depicted in the $V-T$ diagram, we can determine the work done in each segment as follows:
Process $A 	o B$:
This is an isochoric process, meaning that the volume remains constant. In an isochoric process, no work is done because work depends on a change in volume. Therefore, the work done, $W_{AB}$, is:
$   W_{AB} = 0   $
Process $B 	o C$:
This segment is isothermal, meaning the temperature remains constant. For an isothermal process involving an ideal gas, the work done is given by the formula:
$   W_{BC} = nRT_2 \ln \left( \frac{V_1}{V_2} ight)   $
Since we have 1 mole of the gas ($n=1$), the work done simplifies to:
$   W_{BC} = RT_2 \ln \left( \frac{V_1}{V_2} ight)   $
Process $C 	o A$:
This process is isobaric, indicating that the pressure remains constant. In an isobaric process, the work done can be calculated as:
$   W_{CA} = p(V_1 - V_2)   $
Using the ideal gas law, we can equate:
$   p(V_1 - V_2) = nR(T_1 - T_2)   $
Since $n=1$, we further simplify:
$   W_{CA} = R(T_1 - T_2)   $
In summary, the work done in each segment of the process can be stated as:
$W_{AB} = 0$
$W_{BC} = RT_2 \ln \left( \frac{V_1}{V_2} ight)$
$W_{CA} = R(T_1 - T_2)$

Answer

$0, R T_2 \ln \left|\frac{V_1}{V_2}\right|, R\left(T_2-T_1\right)$

Answer

$R\left(T_1-T_2\right), O_1 R T_1 \ln \left|\frac{V_1}{V_2}\right|$

Answer

$0, R T_2 \ln \left|\frac{V_1}{V_2}\right|, R\left(T_1-T_2\right)$

Answer

$0, R T_2 \ln \left|\frac{V_2}{V_1}\right|, R\left(T_2-T_1\right)$

Thermodynamics & KTG

A cyclic process for 1 mole of an ideal is shown in the V-T diagram. The work done in A B, B C and C A respectively is

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