The volume (V) of a monoatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is
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a
b
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d
answer is A.
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Detailed Solution
For the diagram, So we conclude that P = constant.Hence the process AB is isobaric.Now, ΔQ = ΔU + ΔW
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The volume (V) of a monoatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is