- #1

Sweden08

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**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.

Calculate the pressure after the expansion.

**2.**

dU = dQ - dW

dU = dQ - pdV

U = 5/2 NKT

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=?

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=?