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Priyesh (1601)

Blazing goIITian

total posts: 1042
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hi Icy_2291

note: that the temperature will remain constant from a to b as delta U is zero.now it's given that temp at b is 500K. therefore work done from process a to b is nR500ln(2v₀/v₀)_.Now since delta U is zero heat absorbed is equal to work done.similarly from process c to d heat will be released which is equal to nR300ln1/2= -nR300ln2

from process b to c heat absorbed is nc(300-500)= -nc200 where c is the molar heat capacity

from process d to a heat absorbed will be nc200 therefore heat absorbed during b to c & d to a cancel out

therefore total heat absorbed is equal to 2.303nR500log2 + 2.303nR300log1/2

= 2.303nRlog2(200)

= 2.303 * 2 * 8.3 * .3010 * 200 = 2301.433 J 2300 J

"Imagination is more important than knowledge."

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