A cylinder contains oxygen gas (O2) at a pressure of 2.20atm . The volume is 6.0

Question

A cylinder contains oxygen gas (O2) at a pressure of 2.20atm . The volume is 6.00L , and the temperature is 300K . Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes:
(1) Heated at constant pressure from the initial state (state 1) to state 2, which has 420K.
(2) Cooled at constant volume to 250K (state 3).
(3) Compressed at constant temperature to a volume of 6.00L (state 4).
(4) Heated at constant volume to 300K , which takes the system back to state 1.

Calculate Q for each of the four processes.

Calculate W for each of the four processes.

Explanation / Answer

m = ( P1 ) ( V1 ) ( M ) / ( R ) ( T)

=2.2*6*32/0.08205*300

=17.16gm

For heating at constant pressure to 420 K:

Q12 = ( m ) ( CP ) ( T2 - T1 )

Q12 = ( 17.16 g ) ( 0.918 J / g - K ) ( 420 K - 300 K )

Q12 = 1890.34J

V2 = ( V1 ) ( T2 / T1 )

V2 = ( 6.0 L ) ( 420 K / 300 K ) = 8.4 L

W12B = ( P1 ) ( V2 - V1 )

WB12 = ( 2.20 atm ) ( 101325 Pa / atm ) ( 8.4 L - 6.0 L ) ( 1 cu m / 1000 L )

WB12 = 534.996J

For cooling at constant volume to 250 K :
..
Q23 = ( m ) ( CV ) ( T3 - T2 )

Q23 = ( 17.16 g ) ( 0.658 J / g - K ) ( 250 K - 420 K )

Q23 =-1919.51J

WB23 = Integral [ P dV ] = 0.0 J <------------------------

V3 = V2 = 8.4 L

P3 = ( P2 ) ( T3 / T2 )

P3 = ( 2.20 atm ) ( 250 K / 420 K ) = 1.31atm

Now compress at constant T to 6.0 L :
---------------------------------------...

WB34 = ( m / M ) ( R ) ( T3 ) [ ln ( V4 / V3 ) ]

WB34 = ( 17.16g / 32.00 g/gmol ) ( 8.314 J / gmol - K ) ( 250 K ) [ ln ( 6.0 L / 8.4 L ) ]

WB34 = - 375.03 J <---------------------

Delta U34 = ( m ) ( CV ) ( T4 - T3 )

T4 = T3 = 250 K ; therefore Delta U34 = 0.0 J

Q34 = Delta U34 + WB34

Q34 = ( 0.0 J ) + ( - 375.03 J ) = - 375.03 J <----------

Now heat at constant volume back to 300 K :
---------------------------------------...

Q41 = ( m ) ( CV ) ( T1 - T4 )

Q41 = ( 17.16 ) ( 0.658 J / g - K ) ( 300K - 250 K ) = 564.56 J <-----------------------

WB41 = 0.0 J since constant volume <-----------------------------------

For a cycle, you should get: SUM ( Q ) = SUM ( WB )

SUM ( Q ) = Q12 + Q23 + Q34 + Q41

SUM ( Q ) = 1890.34-1919.51-375.03+564.56=160.364J

SUM ( WB ) = WB12 + WB23 + WB34 + WB41

SUM ( WB ) = 534.996+ ( 0.0 ) + ( - 375.03 ) + ( 0.0 ) = 159.966 J

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