0.10 mol of a monatomic gas follows the processshown in the figure Pressure 1 =

Question

0.10 mol of a monatomic gas follows the processshown in the figure
Pressure 1 = (4atm) Volume 1 = (800 cm^3) Pressure 2 = (2atm) Volume 2 = (800cm^3) Pressure 3 = (2atm) Volume 3 = (1600cm^3) a) How much heat energy is transferred to or from ther gasduring the process 1 to 2? b) How much heat energy is transferred to or from the gasduring process 2 to 3? c) What is the total change in thermal energy of thegas?
. 0.10 mol of a monatomic gas follows the processshown in the figure
Pressure 1 = (4atm) Volume 1 = (800 cm^3) Pressure 2 = (2atm) Volume 2 = (800cm^3) Pressure 3 = (2atm) Volume 3 = (1600cm^3) a) How much heat energy is transferred to or from ther gasduring the process 1 to 2? b) How much heat energy is transferred to or from the gasduring process 2 to 3? c) What is the total change in thermal energy of thegas?
.

Explanation / Answer

No.of moles n =0.1 mol Pressure 1 is P = (4atm)= 4 * 101.3*10^3 Pa Volume 1 is V = (800 cm^3)= 800*10^-6 m^3 Pressure 2 is P ' = (2atm)= 2* 101.3 * 10^ 3 Pa Volume 2 is V ' = (800cm^3)= 800* 10 ^ -6 m^3 Pressure 3 is P " = (2atm) = 2 * 101.3 * 10^3 Pa Volume 3 is V " = (1600cm^3)= 1600* 10^-6 m^3 from the relation PV=nRT Temperature 1 is T = PV / nR where R = gas constant = 8.314 J / mol K plug the values we get T = 389.896 K Temperature 2 is T ' = P'V' /nR                               = 194.948 K temperature 3 is T" = P"V"/ nR                              = 389.89 K (a). heat energy is transferred to or from ther gas during theprocess 1 to 2 is Q = nCv ( T ' - T ) where Cv = specific heat at constant volume =1.5* R= 1.5 * 8.314 = 12.471 plug the values we get Q = -243.11 J b) heat energy is transferred to or from the gas duringprocess 2 to 3 is Q ' = n Cp ( T " - T ' ) where Cp = specific heat atconstant pressure =2.5* R = 2.5 * 8.314= 20.785 plug the values weget Q ' = 405.18 J Pressure 1 is P = (4atm)= 4 * 101.3*10^3 Pa Volume 1 is V = (800 cm^3)= 800*10^-6 m^3 Pressure 2 is P ' = (2atm)= 2* 101.3 * 10^ 3 Pa Volume 2 is V ' = (800cm^3)= 800* 10 ^ -6 m^3 Pressure 3 is P " = (2atm) = 2 * 101.3 * 10^3 Pa Volume 3 is V " = (1600cm^3)= 1600* 10^-6 m^3 from the relation PV=nRT Temperature 1 is T = PV / nR where R = gas constant = 8.314 J / mol K plug the values we get T = 389.896 K Temperature 2 is T ' = P'V' /nR                               = 194.948 K temperature 3 is T" = P"V"/ nR                              = 389.89 K (a). heat energy is transferred to or from ther gas during theprocess 1 to 2 is Q = nCv ( T ' - T ) where Cv = specific heat at constant volume =1.5* R= 1.5 * 8.314 = 12.471 plug the values we get Q = -243.11 J b) heat energy is transferred to or from the gas duringprocess 2 to 3 is Q ' = n Cp ( T " - T ' ) where Cp = specific heat atconstant pressure =2.5* R = 2.5 * 8.314= 20.785 plug the values weget Q ' = 405.18 J

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