One mole of an ideal gas is heated at a constant pressure of 1 atm from 00C to 1000 C. Work done by the gas is

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8.31×103J

8.31×10−3J

8.312×10−3J

8.31×102J

answer is D.

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Detailed Solution

dU=dQ−dW⇒dW=dQ−dUFor ideal gas, dU = 0∴ dW=dQ=RT2−T1=831×100=831×102J

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