Isothermal Compression of an Ideal Gas

Example 5-3 Isothermal Compression of an Ideal Gas

A piston-cylinder device initially contains 0.4 m3 of air at 100 kPa and 80C. The air is now compressed to 0.1m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process.

Solution: Air in a piston-cylinder device is compressed isothermally. The boundary work done is to be determined.

Assumption: 1. The compression process is quasi-equilibrium.

2. At specified conditions, air can be considered to be an ideal gas since it is at a high temperature and low pressure relative to its critical-point values.

Analysis: A sketch of the system and the P-v diagram of the process are shown i Fig 5-8. For an ideal gas at constant temperature T_0,

$P V = m R T_{0} = C o r P = \frac{C}{V}$

where C is a constant. Submitting this into Eq. 4-2, we have

$W_{b} = \int_{1}^{2} P d V = C \int_{1}^{2} \frac{d V}{V} = C \ln \frac{V_{2}}{V_{1}} = P_{1} V_{1} \ln \frac{V_{2}}{V_{1}} I n E q, P_{1} V_{1} c a n b e r e p l a c e d b y P_{2} V_{2} o r m R T_{0} . A l s o, \frac{V_{2}}{V_{1}} c a n b e r e p l a c e d b y \frac{P_{1}}{P_{2}} f o r t h i s c a s e \sin c e P_{1} V_{1} = P_{2} V_{2}$

Varför kan man säga $P_{1} V_{1} = P_{2} V_{2}$ ?

För att enligt Figuren 5-8 i boken så ändrade både V, m3 och P i den här processen. Endast temperatur är konstant.

Substituting the numerical values into Eq. 5-7 yields:

$W_{b} = P_{1} V_{1} \ln \frac{V_{2}}{V_{1}} = (100 k P a) (0.4 m 3) (\ln \frac{0.1}{0.4}) = - 55.5 k J$

Man kan använda sig av p₁V₁ = p₂V₂ just fär att temperaturen är konstant. Man kan alltid använda sig av idealgaslagen (under förutsättning att gasen är ideal, förstås): pV = nRT => pV = konstant om T är konstant och substansmängden inte ändras.

Vad betyder "compression process is quasi-equilibrium"?

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