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Following figure shows on adiabatic cylindrical container of volume V subscript 0 end subscript divided by an adiabatic smooth piston (area of cross-section = A) in two equal parts. An ideal gas left parenthesis C subscript P end subscript divided by C subscript V end subscript equals gamma right parenthesis is at pressure P1 and temperature T1 in left part and gas at pressure P2 and temperature T2 in right part. The piston is slowly displaced and released at a position where it can stay in equilibrium. The final pressure of the two parts will be (Suppose x = displacement of the piston)

  1. P subscript 2 end subscript    
  2. P subscript 1 end subscript    
  3. fraction numerator P subscript 1 end subscript open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction close parentheses to the power of gamma end exponent over denominator open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction plus A x close parentheses to the power of gamma end exponent end fraction    
  4. fraction numerator P subscript 2 end subscript open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction close parentheses to the power of gamma end exponent over denominator open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction plus A x close parentheses to the power of gamma end exponent end fraction    

The correct answer is: fraction numerator P subscript 1 end subscript open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction close parentheses to the power of gamma end exponent over denominator open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction plus A x close parentheses to the power of gamma end exponent end fraction


    As finally the piston is in equilibrium, both the gases must be at same pressure P subscript f end subscript. It is given that displacement of piston be in final state x and if A is the area of cross-section of the piston. Hence the final volumes of the left and right part finally can be given by figure as

    V subscript L end subscript equals fraction numerator V subscript 0 end subscript over denominator 2 end fraction plus A x and V subscript R end subscript equals fraction numerator V subscript 0 end subscript over denominator 2 end fraction minus A x
    As it is given that the container walls and the piston are adiabatic in left side and the gas undergoes adiabatic expansion and on the right side the gas undergoes adiabatic compressive. Thus we have for initial and final state of gas on left side
    P subscript 1 end subscript open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction close parentheses to the power of gamma end exponent equals P subscript f end subscript open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction plus A x close parentheses to the power of gamma end exponent.(i)
    Similarly for gas in right side, we have
    P subscript 2 end subscript open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction close parentheses to the power of gamma end exponent equals P subscript f end subscript open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction minus A x close parentheses to the power of gamma end exponent.(ii)
    From eq. (i) and (ii)
    fraction numerator P subscript 1 end subscript over denominator P subscript 2 end subscript end fraction equals fraction numerator open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction plus A x close parentheses to the power of gamma end exponent over denominator open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction minus A x close parentheses to the power of gamma end exponent end fraction Þ A x equals fraction numerator V subscript 0 end subscript over denominator 2 end fraction fraction numerator open square brackets P subscript 1 end subscript superscript 1 divided by gamma end superscript minus P subscript 2 end subscript superscript 1 divided by gamma end superscript close square brackets over denominator open square brackets P subscript 1 end subscript superscript 1 divided by gamma end superscript plus P subscript 2 end subscript superscript 1 divided by gamma end superscript close square brackets end fraction
    Now from equation (i) P subscript f end subscript equals fraction numerator P subscript 1 end subscript open parentheses fraction numerator V subscript 0 end subscript over denominator 2 end fraction close parentheses to the power of gamma end exponent over denominator open square brackets fraction numerator V subscript 0 end subscript over denominator 2 end fraction plus A x close square brackets to the power of gamma end exponent end fraction

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