The internal energy ( U ) of an ideal ( 10.7 mole) gas is plotted against volume for a cyclic process ABCDA , as shown in the figure.
The temperature of the gas at B and C are 500 K and 300 K, respectively. The heat absorbed by the gas (in kcal) in this cyclic process is :

Question

The internal energy ( U ) of an ideal ( 10.7 mole) gas is plotted against volume for a cyclic process ABCDA , as shown in the figure.

The temperature of the gas at B and C are 500 K and 300 K, respectively. The heat absorbed by the gas (in kcal) in this cyclic process is :

TestHub · Accepted Answer

First Law of Thermodynamics: For any cyclic process, the net change in internal energy is $Δ U = 0$ , which means the net heat absorbed equals the total work done ( $Q = W_{n e t}$ ).

Isochoric Paths: The processes $B \to C$ and $D \to A$ occur at a constant volume ( $2 V_{0}$ and $V_{0}$ , respectively), meaning no work is done ( $W_{BC} = W_{D A} = 0$ ).

Isothermal Paths: The paths $A \to B$ and $C \to D$ are constant internal energy processes (constant $U ⟹$ constant $T$ ) at temperatures $T_{B} = 500 K$ and $T_{C} = 300 K$ .

Work Calculation: The net work is the sum of the isothermal steps:

$W_{n e t} = n R T_{B} ln (\frac{2 V _{0}}{V _{0}}) + n R T_{C} ln (\frac{V _{0}}{2 V _{0}}) = n R ln (2) (T_{B} - T_{C})$

Final Result: Substituting $n = 10.7 m o l$ , $R \approx 2 c a l / m o l \cdot K$ , and $ln (2) \approx 0.693$ , we get $Q = 10.7 \times 2 \times 0.693 \times (500 - 300) \approx 2966 c a l = 2.97 \approx 3.00kcal$ .

Chemistry - Thermodynamics - II Question with Solution | TestHub

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