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One mole of a gas expands with temperature T such thaht its volume, V=`KT^(2)`, where K is a constant. If the temperature of the gas changes by `60^(@)C` then the work done by the gas is `120RA. R ln 60B. kR In 60C. 60 kRD. 120R |
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Answer» Correct Answer - D `V=kT^(2) therefore 2kT dT=dV` and `P=(RT)/(V)=(RT)/(kT^(2))=(R)/(kT)` `dW=PdV=2RdT, W=int PdV` |
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