JEE MAIN - Chemistry (2018 - 15th April Morning Slot - No. 10)
An ideal gas undergoes a cyclic process as shown in Figure.
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$$\Delta $$UBC = $$-$$5 kJ mol-1, qAB = $$2$$ kJ mol-1, WAB = $$-$$5 kJ mol-1, WCA = 3 kJ mol-1. Heat absorbed by the system during process $$CA$$ is :
$$\Delta $$UBC = $$-$$5 kJ mol-1, qAB = $$2$$ kJ mol-1, WAB = $$-$$5 kJ mol-1, WCA = 3 kJ mol-1. Heat absorbed by the system during process $$CA$$ is :
$$-$$5 kJ mol-1
+5 kJ mol-1
18 kJ mol-1
$$-$$18 kJ mol-1
Explanation
Given,
$$\Delta $$UBC = $$-$$ 5 kJ/mol
$$\Delta $$UAB = qAB + WAB
= 2 + ($$-$$ 5)
= $$-$$ 3 kJ/mol.
For cyclic process,
$$\Delta $$U = 0
$$ \Rightarrow $$$$\,\,\,\,$$ $$\Delta $$UAB + $$\Delta $$UBC + $$\Delta $$UCA = 0
$$ \Rightarrow $$$$\,\,\,\,$$ $$\Delta $$UCA = $$-$$($$-$$3) $$-$$ ($$-$$ 5) = 8 kJ/mol
As, $$\,\,\,\,\,\,$$ $$\Delta $$UCA = qCA + WCA
$$ \Rightarrow $$$$\,\,\,\,$$ 8 = qCA + 3
$$ \Rightarrow $$$$\,\,\,\,$$ qCA = + 5 kJ/mol.
$$\Delta $$UBC = $$-$$ 5 kJ/mol
$$\Delta $$UAB = qAB + WAB
= 2 + ($$-$$ 5)
= $$-$$ 3 kJ/mol.
For cyclic process,
$$\Delta $$U = 0
$$ \Rightarrow $$$$\,\,\,\,$$ $$\Delta $$UAB + $$\Delta $$UBC + $$\Delta $$UCA = 0
$$ \Rightarrow $$$$\,\,\,\,$$ $$\Delta $$UCA = $$-$$($$-$$3) $$-$$ ($$-$$ 5) = 8 kJ/mol
As, $$\,\,\,\,\,\,$$ $$\Delta $$UCA = qCA + WCA
$$ \Rightarrow $$$$\,\,\,\,$$ 8 = qCA + 3
$$ \Rightarrow $$$$\,\,\,\,$$ qCA = + 5 kJ/mol.
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