An 1.2 m3 rigid tank contains nitrogen gas at 200 kPa and 350 k Now heat is tran

Question

An 1.2 m3 rigid tank contains nitrogen gas at 200 kPa and 350 k Now heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 900 kPa. The work done during this process is 0 kJ 1080 kJ 840 kJ 156 kJ 240 kJ A cylinder contains 2 kg of steam at 200 kPa and 200 C. Now steam is compressed at constant pressure until its temperature rises to 250 C. The work done during this compression process is W = m * w W_b = integral_1^2 PdV W_b = P Delta V Isobaric W_b = P_2 V_2 - P_1 V_1/1 - n W_b = P_1 V_1 In V_1/V_1 = m RT In V_2/V_1 Isothermal

Explanation / Answer

Given

   nitrogen gas initially v1 = 1.2 m^3 , P1 = 200 kPa, T1 = 350 k

form ideal gas equation PV = nRT ==> n = PV/RT

               n = 200*10^3*1.2/(8.314*350)
               n = 82.5

now heat has been supplied and the pressure changed to P2 = 900 kPa

           work done W = 2.303*nRTln(p1/P2)

               W = 2.303*82.5*8.414*350*ln(200/900)

               W = 841 kJ ---------------->>> answer

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