Internal Energy (E,also denoted by U): Every system having some quantity of matter is associated with a definite amount of energy, called internal energy. E=E_("Transiational")+E_("Rotational")+E_("Vibrational")+E_("bonding")+.... DeltaE=E_("Final")-E_("initial") DeltaE=q_v, heat supplied to a gas at constant volume, since all the heat supplied goes to increase the internal energy to the gas. It is an extensive property & a state function.It is exclusively a function of temperature. If DeltaT=0 , DeltaE=0 as well. The internal energy of a certain substance is given by the following equation : U=3 PV+84 where U is given in kJ/kg, P is in kPa, and V is in m^3//kg A system composed of 3 kg of this substance expands from an initial pressure of 400 kPa and a volume of A 0.2 m^3 to a final pressure 100 kPa in a process in which pressure and volume are related by PV^2=constant. If the expansion is quasi-static, then the value of q is :

Internal Energy (E,also denoted by U): Every system having some quant

1.	Internal Energy (E,also denoted by U): Every system having some quantity of matter is associated with a definite amount of energy, called internal energy. E=E_("Transiational")+E_("Rotational")+E_("Vibrational")+E_("bonding")+.... DeltaE=E_("Final")-E_("initial") DeltaE=q_v, heat supplied to a gas at constant volume, since all the heat supplied goes to increase the internal energy to the gas. It is an extensive property & a state function.It is exclusively a function of temperature. If DeltaT=0 , DeltaE=0 as well. The internal energy of a certain substance is given by the following equation : U=3 PV+84 where U is given in kJ/kg, P is in kPa, and V is in m^3//kg A system composed of 3 kg of this substance expands from an initial pressure of 400 kPa and a volume of A 0.2 m^3 to a final pressure 100 kPa in a process in which pressure and volume are related by PV^2=constant. If the expansion is quasi-static, then the value of q is :
Answer» 80 kJ 60 kJ 40 kJ 120 kJ Solution :U=3PV+84 `DeltaU=U_2-U_1=3(P_2V_2-P_1V_1)` `THEREFORE DeltaU=3(P_2V_2-P_1V_1)` Now, `P_1V_1^2=P_2V_2^2` `:. V_2=V_1(P_1/P_2)^(1//2)=0.2(4/1)^(1//2)=0.4` `DeltaU=300(1xx0.4-4xx0.2)kJ = -120 kJ` For a quasi-static process `W=intPdV=(P_2V_2-P_1V_1)/(n-1)` `=((1xx0.4-4xx0.2)100)/(2-1)=-40 kJ` `:. DeltaU=q+W` q=-120+40=-80 kJ