I am not even sure will to begin with this problem any suggestions? Two cylinder

Question

I am not even sure will to begin with this problem any suggestions? Two cylinders linked by a narrow tube with a valve, V. Initially V is evacuated. A cylinder, A having VA = 30 liter, is connected to a second cylinder, B, of circular cross-section, as shown in Fig. 3.16. Cylinder B is fitted with a piston of diameter 10cm, which without friction or leaking. Initially A contains an ideal nonatomic gas; in addition, PA = 10 MPa, TA = 300 K, VB=0 and the compression force on the spring is zero. The spring, which Hooke's law, has a force constant of 1 times 10 5 Nm-1. The valve, V, is opened slowly and the gas leaks through from A to B, compressing the spring. Equilibrium is eventually established again at a temperature of 300 K throughout all parts of the system. Assume the pressure on the right-hand side of the piston can be neglected; neglect the volume of the connecting pipe; and assume the heat capacity of the cylinders can be ignored. Calculate the distance by which the spring is compressed in the final equilibrium state. Determine the work done on the gas. calculate the total heat transferred to the gas during this process. Is the process described quasistatic or not? Consider again the system shown in Fig. 3.16. The initial state of the apparatus is described in question 3.6.3. An externally applied cylinder of fixed volume (A) coupled to a cylinder fitted with a piston Initially the valve V is closed, the volume of B is zero, and the com due to the spring is zero. The volume of the connecting tube is

Explanation / Answer

The initial volume of tube A is VA = 30 Lt = 30*10^-3 m^3 The initial pressure of the gas is PA = 10 MPa = 10^7 Pa The initial temperature of the gas is TA = 300 K After the valve is opened, The final volume is V' = 30 Lt + 10 Lt = 40 Lt = 40*10^-3 m^3 The final temperature of the gas T' = 300 K The Idealgas equation: PAVA/TA = P'V'/T' P' = PAVAT'/V'TA = (10^7 Pa)(30*10^-3 m^3)(300 K)/(40*10^-3 m^3)(300 K) = 7.5*10^6 Pa This is the pressure of the gas at the equilibrium condition The pressure acting on the piston is P = 7.5*10^6 Pa The force acting on the piston is F = PA = (7.5*10^6 Pa)(3.14*0.05*0.05) = 5.8875*10^3 N Let x be the compression in the spring, then According to Hooke's law F = kx x = F/k = 5.8875*10^3 N/10^5 Nm^-1 = 5.8875 cm The workdone on the gas is given by W = P'(V'-V0)+0.5*(P0-P')(V'-V0) = (7.5*10^6)*(10*10^-3)+0.5*(2.5*10^6)*(10*10^-3) = -87.5*10^3 J (work is done on the gas) As the process is isothermal, the change in internal energy of the system is zero The heat transferred will be equal to the workdone DELTA Q = DELTA U + W DELTA Q = 87.5*10^3 J As the temperature of the system is not changed, the process should be the Quasi-static process

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