JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 8)
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the
internal energy of the gas along the path ca is –180 J. The gas absorbs 250 J of heat along the path ab and 60
J along the path bc. The work done by the gas along the path abc is:
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120 J
130 J
100 J
140 J
Explanation
For the process (c – a), $$\Delta $$Uca = – 180 J
For process (b – c) $$ \to $$ Isochoric (Wbc = 0)
$$ \therefore \Delta $$U = 60 J
Heat absorbs along (a – b), Qab = 250 J
Also $$ \therefore \Delta$$ Ucycle = 0
$$ \therefore \Delta$$ Uab = 120 J
So Wa $$ \to $$ b = 130 J
Total work done from (a $$ \to $$ b $$ \to $$ c)
= Wab + Wbc = 130 J
For process (b – c) $$ \to $$ Isochoric (Wbc = 0)
$$ \therefore \Delta $$U = 60 J
Heat absorbs along (a – b), Qab = 250 J
Also $$ \therefore \Delta$$ Ucycle = 0
$$ \therefore \Delta$$ Uab = 120 J
So Wa $$ \to $$ b = 130 J
Total work done from (a $$ \to $$ b $$ \to $$ c)
= Wab + Wbc = 130 J
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