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A quantity of 5. 00 mol of a monatomic ideal gas has an initial pressure of pa=1. 00×105pa and an initial volume of va=0. 0200m3. Two paths are used to take the gas to a final state for which pb=1. 80×105pa and vb=0. 0280m3

Sagot :

The total work done ( W ) = 1440 J  

The total heat ( Q ) = 6000 J

Given data :

Initial pressure ( Pa ) = 1 * 10⁵ pa

final pressure ( Pb ) = 1.8 * 10⁵ pa

Determine the Total work done and Heat

Considering process A --> B ( iso -choric process )

Wa -->b = 0

ΔQa --->b = nCvdT ---- ( 1 )

Note : dT = vdP / nR

equation ( 1 ) becomes

ΔQa -->b = nCv (VadP / nR )

               = n( 3/2R ) [ va ( pb - pa ) / nR ]

               = 3/2 * ( 0.0200 ) * ( 1.80 - 1 ) * 10⁵ pa

     Δ Qab  = 2400 J

Considering process B ---> C

Wb --> c = pdv

              = pb ( vb - va )

Wb --> c = 1.80 * 10⁵ ( 0.0280 - 0.0200 )m³

          Wbc  = 1440 J

ΔQbc  = nCp dT

           = n(5/2 R ) ( pdv/nR )

           = 5/2 [ Pb (Vb -Va) ]

           = 5/2 * 1440

 ΔQbc = 3600 J

Therefore :

The Total work done( W ) = Wab + Wbc

                                          = 1440 J

Total Heat ( Q ) =  ΔQab +  ΔQbc

                         = 2400 + 3600 = 6000 J

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