We want to raise the temperature of 3.50 moles of an ideal gas from -20.0 degree

Question

We want to raise the temperature of 3.50 moles of an ideal gas from -20.0 degrees celsius to 89.0 degrees celsius.

a. Calculate the amount of heat (in joules) needed if the gas is ideal argon at a fixed volume of 8.80 meters cubed.

b.Calculate the amount of heat (in joules) needed if the gas is ideal helium at a constant pressure of 1.07 atm.

c. Calculate the amount of heat (in joules) needed if the gas is ideal nitrogen at a constant pressure of 1.26 atm.

d. Calculate the amount of heat (in joules) needed if the gas is ideal chlorine at a constant volume of 24.6 L.

Explanation / Answer

According to first   law of thermodynamics                     Q = dU + W (a)    at   constant   volume   dV   =0    hence   W = PdV = 0 heat required   Q = dU    = nCvdT Here n = 3.5 moles   , Cv   =   (3/2) R   -- for monoatomis gas change in tepmperature   dT = 109 K   , R = 8.314 J/mole.K then    Q   = 4757.7 J -------------------------------------------------- -------------------------------------------------- (b)                 Q = dU + W    work   done   W = PdV =   nRdT   = (3.5 )(8.314 )( 109) =3171.8 J change ininternal energy at   constant pressure  for helium = n CpdT =    Cp   = specific heat at constant pressure   = 20.8 J /mole .K    dU      = (3.5 ) ( 20.8 J /mole .K ) (109) = 7935.2 J         Q = 3171.8 J + 7935.2 J   = 11107 J -------------------------------------------------------- (c) For nitrogen ( diatomic gas )    Cp   = 29.1 J/mole .K    Q = dU + W    work   done   W = PdV =   nRdT   = (3.5 )(8.314 )( 109) =3171.8 J change ininternal energy at   constant pressure  for helium = n CpdT =       dU      = (3.5 ) ( 29.1 J /mole .K ) (109) =11101.65 J                 Q = 14273.45 J -------------------------------------------------------- (d)   For chlourin gas   (diatomic gas )   Cv   = 25.7 J/mole.K at   constant   volume   dV   =0    hence   W = PdV = 0 heat required   Q = dU    = nCvdT Here n = 3.5 moles   , Cv   = 25.7 J/mole.K      -- for monoatomis gas change in tepmperature   dT = 109 K   , R = 8.314 J/mole.K then    Q   =  9804.55 J          change ininternal energy at   constant pressure  for helium = n CpdT =    Cp   = specific heat at constant pressure   = 20.8 J /mole .K    dU      = (3.5 ) ( 20.8 J /mole .K ) (109) = 7935.2 J         Q = 3171.8 J + 7935.2 J   = 11107 J -------------------------------------------------------- (c) For nitrogen ( diatomic gas )    Cp   = 29.1 J/mole .K    Q = dU + W    work   done   W = PdV =   nRdT   = (3.5 )(8.314 )( 109) =3171.8 J change ininternal energy at   constant pressure  for helium = n CpdT =       dU      = (3.5 ) ( 29.1 J /mole .K ) (109) =11101.65 J                 Q = 14273.45 J -------------------------------------------------------- (d)   For chlourin gas   (diatomic gas )   Cv   = 25.7 J/mole.K at   constant   volume   dV   =0    hence   W = PdV = 0 heat required   Q = dU    = nCvdT Here n = 3.5 moles   , Cv   = 25.7 J/mole.K      -- for monoatomis gas change in tepmperature   dT = 109 K   , R = 8.314 J/mole.K then    Q   =  9804.55 J          change ininternal energy at   constant pressure  for helium = n CpdT =       dU      = (3.5 ) ( 29.1 J /mole .K ) (109) =11101.65 J                 Q = 14273.45 J -------------------------------------------------------- (d)   For chlourin gas   (diatomic gas )   Cv   = 25.7 J/mole.K at   constant   volume   dV   =0    hence   W = PdV = 0 heat required   Q = dU    = nCvdT Here n = 3.5 moles   , Cv   = 25.7 J/mole.K      -- for monoatomis gas change in tepmperature   dT = 109 K   , R = 8.314 J/mole.K then    Q   =  9804.55 J          change ininternal energy at   constant pressure  for helium = n CpdT =       dU      = (3.5 ) ( 29.1 J /mole .K ) (109) =11101.65 J                 Q = 14273.45 J -------------------------------------------------------- (d)   For chlourin gas   (diatomic gas )   Cv   = 25.7 J/mole.K at   constant   volume   dV   =0    hence   W = PdV = 0 heat required   Q = dU    = nCvdT Here n = 3.5 moles   , Cv   = 25.7 J/mole.K      -- for monoatomis gas change in tepmperature   dT = 109 K   , R = 8.314 J/mole.K then    Q   =  9804.55 J          heat required   Q = dU    = nCvdT Here n = 3.5 moles   , Cv   = 25.7 J/mole.K      -- for monoatomis gas change in tepmperature   dT = 109 K   , R = 8.314 J/mole.K then    Q   =  9804.55 J         

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