Four kilograms of an ideal gas are contained within a piston-cylinder device. The gas undergoes a polytropic process with a polytropic exponent of 1.5. The initial pressure is 3 bars, the initial volume is 0.1 m3, and the final volume is 0.2 m3. If the change in specific internal energy of the gas during the process is -4.6 kJ/kg, determine the net heat transfer, in kJ.

The net heat transfer is -0.4 kJ.

Explanation:

For a polytropic process,

P1V1^n = P2V2^n

P1 (initial pressure) = 3 bars = 3×100 = 300 kPa

V1 (initial volume) = 0.1 m^3

V2 (final volume) = 0.2 m^3

n (polytropic exponent) = 1.5

P2 = P1(V1/V2)^n = 300(0.1/0.2)^1.5 = 300(0.5)^1.5 = 300 × 0.35 = 105 kPa

W (work done) = (P1V1 - P2V2)/n - 1 = (300×0.1 - 105×0.2)/1.5 - 1 = (30 - 21)/0.5 = 9/0.5 = 18 kJ

∆U (change in internal energy) = change in specific internal energy × mass = -4.6 kJ/kg × 4 kg = -18.4 kJ

Q (net heat transfer) = ∆U + W = -18.4 kJ + 18 kJ = -0.4 kJ

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

answerca air mass;