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Figure below shows two paths that may be taken by a gas to go from a state $A$ to a 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 ...... $J$
$380 $
$500 $
$460$
$300 $
Solution
As initial and final points are same so
$\Delta {U_{ABC}} = \Delta {U_{AC}}$
$AB$ is isochoric process.
$\Delta {W_{AB}} = 0$
$\Delta {Q_{AB}} = \Delta {U_{AB}} = 400\,J$
$BC$ is isobaric process.
$\Delta {Q_{BC}} = \Delta {U_{BC}} + \Delta {W_{BC}}$
$100 = \Delta {U_{BC}} + 6 \times {10^4}\left( {4 \times {{10}^{ – 3}} – 2 \times {{10}^{ – 3}}} \right)$
$100 = \Delta {U_{BC}} + 12 \times 10$
$\Delta {U_{BC}} = 100 – 120 = – 20\,J$
$As,\,\Delta {U_{ABC}} = \Delta {U_{AC}}$
$\Delta {U_{AB}} + \Delta {U_{BC}} = \Delta {Q_{AC}} – \Delta {W_{AC}}$
$400 – 20 = \Delta {Q_{AC}} – (2 \times {10^4} \times 2 \times {10^{ – 3}} + \frac{1}{2} \times $
$2 \times {10^{ – 3}} \times 4 \times {10^4})$
$\Delta {Q_{AC}} = 460J$
