- #1
Sweden08
- 1
- 0
1.1 kg off air with pressure 10^6 Pa, temperature 125 celciusdegrees, gets to expand until the volume is five times bigger than in the beginning. In every moment of the expansion, the amount of (warmth energy = Q) contributed is one fourth of the work done by the gas. 1 kmol of air weighs about 29 kg, Cv=5/2R.
Calculate the pressure after the expansion.
2.
dU = dQ - dW
dU = dQ - pdV
U = 5/2 NKT
3. Since this process is not konstant for either temperature, volume, pressure, i have no clue how to solve it. However:
If dQ = 1/4 dW, then dU = 1/4 dW - dW = -3/4 dW
And since U = 5/2 NKT = 5/2 PV. ---> dU = 5/2 (P1V1-P0V0)
P0= start pressure, V0=start volume, V1= endvolume = 5V0, P1=endpressure=?
Calculate the pressure after the expansion.
2.
dU = dQ - dW
dU = dQ - pdV
U = 5/2 NKT
3. Since this process is not konstant for either temperature, volume, pressure, i have no clue how to solve it. However:
If dQ = 1/4 dW, then dU = 1/4 dW - dW = -3/4 dW
And since U = 5/2 NKT = 5/2 PV. ---> dU = 5/2 (P1V1-P0V0)
P0= start pressure, V0=start volume, V1= endvolume = 5V0, P1=endpressure=?