Question 23: Thermodynamics & KTG

Question

In the given cycle $ A B C D A $, the heat required for an ideal monoatomic gas will be
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Accepted Answer

In the cycle $ A \rightarrow B \rightarrow C \rightarrow D \rightarrow A $, heat is added to the system during the processes $ D \rightarrow A $ (at constant volume) and $ A \rightarrow B $ (at constant pressure).
Calculating the Heat Added
Heat for Process $ D \rightarrow A $ (Constant Volume):
For an ideal monoatomic gas, the molar heat capacity at constant volume $ C_V $ is given by:
$    C_V = \frac{3}{2} R    $
The heat added during this process is:
$    Q_{DA} = n C_V (\Delta T)_{DA} = \frac{3}{2} n R \Delta T    $
Heat for Process $ A \rightarrow B $ (Constant Pressure):
The molar heat capacity at constant pressure $ C_p $ is given by:
$    C_p = \frac{5}{2} R    $
The heat added during this process is:
$    Q_{AB} = n C_p (\Delta T)_{AB} = \frac{5}{2} n R \Delta T    $
Total Heat Supplied
The total heat supplied in the cycle is the sum of the heat added during these two processes:
$ Q = Q_{DA} + Q_{AB} $
Substitute the expressions for $ Q_{DA} $ and $ Q_{AB} $:
$ Q = \frac{3}{2} n R \Delta T + \frac{5}{2} n R \Delta T $
Given $ n R \Delta T = p \Delta V $, and substituting it:
$ Q = \frac{3}{2}(p_0 V_0) + \frac{5}{2}(p_0 V_0) $
This simplifies to:
$ Q = \frac{13}{2} p_0 V_0 $

Answer

$ p_{0} V_{0} $

Answer

$ \frac{13}{2} p_{0} V_{0} $

Answer

$ \frac{11}{2} p_{0} V_{0} $

Answer

$ 4 p_{0} V_{0} $

Thermodynamics & KTG

In the given cycle A B C D A , the heat required for an ideal monoatomic gas will be

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