A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where the pressure is 1.0 bar, during which the pressure–volume relationship is pV = constant. Assume ideal gas behavior for the air.

Determine the work and heat transfer, in kJ.

For an ideal gas, the specific volume is equal to:

A: the gas constant R times the gage temperature divided by the gage pressure.

B: the universal gas constant times the absolute temperature divided by the gage pressure.

C: the gas constant R times the absolute temperature divided by the absolute pressure.

D: the universal gas constant times the absolute temperature divided by the absolute pressure.

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Xema8 Xema8 · Answer 1 · 2020-03-29T10:42:02+02:00

Answer:

Explanation:

The process is isothermic, as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303 log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv = m / M RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.