Question
An ideal gas goes from state A to state B via three different processes as indicated in the p-V diagram.
If Q1, Q2, Q3 indicate the heat absorbed by the gas along the three processes and ΔU1, ΔU2, ΔU3 indicate the change in internal energy along the three processes respectively, then
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Q1> Q2> Q3 and ΔU1=ΔU2= ΔU3
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Q3> Q2> Q1 and ΔU1=ΔU2= ΔU3
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Q1= Q2= Q3 and ΔU1=ΔU2= ΔU3
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Q3> Q2> Q1 and ΔU1>ΔU2> ΔU3
Solution
A.
Q1> Q2> Q3 and ΔU1=ΔU2= ΔU3
For all process 1, 2 and 3
ΔU = UB -UA is same
Therefore, ΔU1 = ΔU2 = ΔU3
Now, ΔQ = ΔU+ΔW
Now, ΔW = work done by the gas
therefore
ΔQ1 >ΔQ2> ΔQ3
ΔQ1 >ΔQ2> ΔQ3
