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Two systems have the following equations of state and are contained in a closed cylinder, separated by a fixed, adiabatic and impermeable piston. N1= 2 and N2 = 1.5 moles. The initial temperatures are T1 = 175 K and T2 = 400 K. The total volume is 0.025 m^3. The piston is allowed to move and heat transfer is allowed across the piston. Determine the final temperature of the system (in Kelvin). 1/T1=(3/2)R*N1/U1 ,P1/T1=R*N1/V1 and 1/T2=(5/2)R*N2/U2 , P2/T2=R*N2/V2 |
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Answer» 1= 2 and N2 = 1.5 moles. The initial temperatures are T1 = 175 K and T2 = 400 K. The total volume is 0.025 m^3. 1/T1=(3/2)R*N1/U1 ,P1/T1=R*N1/V11/T2=(5/2)R*N2/U2 , P2/T2=R*N2/V2To find:Determine the final temperature of the system (in Kelvin).Solution:Let the volumes of two systems be, V1 and V2From given, we have,V1 + V2 = V = 0.025 m³As given, that both systems are in a closed CONTAINER in ADIABATIC condition, we have,T1 = 175 K T2 = 400 Kw.k.t PV = nRTnRT1/P1 + nRT2/P2 = nRT/P2R×175/P + 1.5R×400/P = 3.5RT/P2 × 175 + 1.5 × 400 = 3.5T350 + 600 = 3.5T950 = 3.5TT = 950/3.5∴T = 271.42 KHence the final temeperature. |
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