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The volume of 1 kg of helium in a pistonā€“cylinder device is initially 5 m3. Now helium is compressed to 2 m3 while its pressure is maintained constant at 174 kPa. Determine the initial and final temperatures of helium as well as the work required to compress it, in kJ. The gas constant of helium is given to be R = 2.0769 kJ/kgĀ·K. (Round the temperatures to one decimal place and the work required to the nearest whole number.)

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Answer:

Tā‚ = 418.9 k = 145.9Ā° C

Tā‚‚ = 167.6 k = - 105.4Ā° C

W = - 522 KJ

Here negative sign indicates that work is being done on the gas.

Explanation:

For the temperature of Helium, we use the general gas equation:

PV = mRT

For initial state:

PVā‚ = mRTā‚ Ā  Ā (Constant Pressure)

where,

P = constant pressure = 174 KPa

Vā‚ = Initial Volume = 5 mĀ³

m = mass = 1 kg

R = Gas Constant of Helium = 2.0769 KJ/kg.k

Tā‚ = initial temperature of helium = ?

Therefore,

(174 KPa)(5 mĀ³) = (1 kg)(2.0769 KJ/kg.k)Tā‚

Tā‚ = (870 KJ)/(2.0769 KJ/k)

Tā‚ = 418.9 k = 145.9Ā° C

For final state:

PVā‚‚ = mRTā‚‚ Ā  Ā (Constant Pressure)

where,

P = constant pressure = 174 KPa

Vā‚‚ = Final Volume = 2 mĀ³

m = mass = 1 kg

R = Gas Constant of Helium = 2.0769 KJ/kg.k

Tā‚‚ = Final temperature of helium = ?

Therefore,

(174 KPa)(2 mĀ³) = (1 kg)(2.0769 KJ/kg.k)Tā‚‚

Tā‚‚ = (348 KJ)/(2.0769 KJ/k)

Tā‚‚ = 167.6 k = - 105.4Ā° C

The work done on a gas in a constant pressure process (isobaric) is given as:

W = P(Vā‚‚ - Vā‚)

W = (174 KPa)(5 mĀ³ - 2 mĀ³)

W = - 522 KJ

Here negative sign indicates that work is being done on the gas.

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