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Current Question (ID: 11060)
Question:
$\text{The figure below shows two paths that may be taken by a gas to go from state A to state C. In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be:}$ $\text{Given P-V diagram with coordinates: A}(2 \times 10^{-3} \text{ m}^3, 2 \times 10^4 \text{ Pa})\text{, B}(2 \times 10^{-3} \text{ m}^3, 6 \times 10^4 \text{ Pa})\text{, C}(4 \times 10^{-3} \text{ m}^3, 6 \times 10^4 \text{ Pa})$
Options:
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1. $380 \text{ J}$
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2. $500 \text{ J}$
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3. $460 \text{ J}$
(Correct)
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4. $300 \text{ J}$
Solution:
$\text{Since the initial and final points are the same:}$ $\Delta U_{A \rightarrow B \rightarrow C} = \Delta U_{A \rightarrow C} \text{ ......(i)}$ $\text{Also A } \rightarrow \text{ B is isochoric process}$ $\text{So } \Delta W_{A \rightarrow B} = 0$ $\text{and } \Delta Q = \Delta U + \Delta W$ $\text{So, } \Delta Q_{A \rightarrow B} = \Delta U_{A \rightarrow B} = 400 \text{ J}$ $\text{Next B } \rightarrow \text{ C is isobaric process.}$ $\text{So, } \Delta Q_{B \rightarrow C} = \Delta U_{B \rightarrow C} + \Delta W_{B \rightarrow C}$ $= \Delta U_{B \rightarrow C} + P \Delta V_{B \rightarrow C}$ $\Rightarrow 100 = \Delta U_{B \rightarrow C} + 6 \times 10^4 (2 \times 10^{-3})$ $\Rightarrow \Delta U_{B \rightarrow C} = 100 - 120 = -20 \text{ J}$ $\text{From Equation (i):}$ $\Delta U_{A \rightarrow B} + \Delta U_{B \rightarrow C} = \Delta Q_{A \rightarrow C} - \Delta W_{A \rightarrow C}$ $\Rightarrow 400 + (-20) = \Delta Q_{A \rightarrow C} - \text{Area under AC}$ $\Delta Q_{A \rightarrow C} = 380 + (2 \times 10^4 \times 2 \times 10^{-3} + \frac{1}{2} \times 2 \times 10^{-3} \times 4 \times 10^4)$ $= 380 + (40 + 40)$ $\Delta Q_{A \rightarrow C} = 460 \text{ J}$
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