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Air in a 0.3-m3 cylinder is initially at a pressure of 10 bar and a temperature of 330 K. The cylinder is to 3 be emptied by opening a valve and letting the pres- sure drop to that of the atmosphere. What will be the temperature and mass of gas in the cylinder if this is accomplished? a. In a manner that maintains the temperature of the gas at 330 K? b. In a well-insulated cylinder? For simplicity assume, in part (b), that the process oc- curs suficiently rapidly that there is no heat transfer between the cylinder walls and the gas. The gas is ideal, and Cp 29 (J/mol K)

Respuesta :

Answer:

Explanation:

Since temperature is to be maintained ; constant temperature and a such, the process is ISOTHERMAL

  • T1 = T2 = 330k
  • P1 = 10bar
  • to calculate mass of then gas = P1V/RT1
  • m1 = 10 x 10^5 x 0.3 / 287 x 330
  • = 3.1676kg

  • similarly for m2 = P2V/RT2
  • = 10^5 x 0.3 /287 x 330
  • = 0.31676kg
  • hence required mass = m1 - m2
  • = 2.85kg

b) Considering a well-insulated cylinder ;

Cp = 29J/molK

Temperature T2 = T1 [ P1/P2]^(1-γ)/γ

where γ = 1.4

plugging the values into the equation ; T2 = 171k

hence mass of the gas m1 = P1V/RT1

= 10 x 10^5 x 0.3/287 x 171

= 6.11kg

mass of m2 = P2V/RT2

= 10^5 x 0.3/287 x 171

= 0.611kg

hence required mass = m1 - m2

= 5.50kg

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