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tensa zangetsu [6.8K]
3 years ago
7

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. 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.
Physics
1 answer:
vladimir2022 [97]3 years ago
4 0

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table.  R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ ln[\frac{P_{1} }{P_{2} }]

   = <em>mRT</em>₁ln[\frac{P_{1} }{P_{2} }]      [∵<em>pV</em> = <em>mRT</em>]

   = (5) (0.287) (272.039) ln[\frac{2.0}{1.0}]

   = 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = mc_{v}(T_{2}  - T_{1} ) = 0

From 1st law of thermodynamics

Q = Δ<em>U  </em>+ <em>W</em>

   = 0 + 270.588

   = 270.588 kJ

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