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Two moles of a certain ideal gas at temperature T_0= 300 K were cooled isochorically so that the gas pressure reduced eta= 2.0 times. Then, as a result of the isobaric process, the gas expanded till its temperature get back to the initial value. Find the total amount of heat absorbed by the gas in this process. |
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Answer» Solution :In the first process, under isochoric process W= 0 (as `DeltaV=0`). From gas law, IFTHE pressure is reduced to `eta` times, then the temperature is ALSO reduced to `eta` times i.e., the new temperature becomes `T_0//eta`. Thus from firstlaw of thermodynamics, we have `Q_1=DeltaU_1=n C_V DeltaT=(nR)/(gamma-1)DeltaT` or `Q_1=(nR)/(gamma-1)[T_0/eta-T_0]=(nRT_0(1-eta))/(eta(gamma-1))` During second process (under ISOBARIC process), workdone is equal to `PDeltaV=nRDeltaT`. And from first law of thermodynamics, we have `Q_2=DeltaU_2+W_2 =(nRDeltaT)/(gamma-1)+nR DeltaT` `=nRDeltaT[1/(gamma-1)+1]=nR DeltaT[gamma/(gamma-1)]` `=(nRgamma)/(gamma-1)[T_0-T_0/eta]=(nRgamma(eta-1))/(eta(gamma-1))` Now total amount ofheat supplied is `Q=Q_1+Q_2` `=(nRT_0(1-eta))/(eta(gamma-1))+(nRgamma(eta-1))/(eta(gamma-1))` `=nRT_0 (1-1/eta)` Here we haven=2, R=8.3, `T_0` =300 K and `eta=2` Thus `Q=2xx8.3xx300(1-1/2)J` or =2490 J = 2.5 kJ |
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