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Question

One mole of an ideal gas (γ=1.4) at 500 K, is filled in an adiabatic cylinder with a piston which is free to move against atmospheric pressure as shown. If a non-conducting fan is inserted into the gas space and rotated virgorously, so that the gas expands slowly till its volume gets doubled. If the work done by the fan is 7x kJ, then value of x is . Write upto two digits after the decimal point.
(Take R = 8 J/mole.K, neglect heat capacity of the cylinder, piston and the fan).


Solution

Using 1st law of Thermodynamics
Q=W+U0=((Wfan)+PΔV)+nf2RΔT
Wfan=nRΔT+nf2RΔT
Wfan=nCpΔT.

We know Cp=Cv+R
Cv=f2R and as γ=1.4, degrees of freedom f is 5.

So, as the gas is expanding slowly means P is constant. So TV and as the volume is doubled, the temperature will also be doubled and ΔT will be T.

Wfan=(1)(R+f2R)(500 K)   
Wfan=(1)(8+528)(500) kJ 
Wfan=14 kJ

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