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A frictionless piston-cylinder device contains 2 kg of argon at 200 kPa and 0C. Argon is now compressed according to the relation Pv = constant, with = 1:4 until it reaches a final temperature of 80C. Model the system as an ideal gas and calculate the work input during this process.

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olatunjiolorundare20

Answer:-3,325.6J or -3.326Kj

Explanation:

The work done on a gas by a frictionless piston is given by

Work done =∆pv=P1V1-P2V2

Given an ideal gas situation

We have the relationship

PV=nRT where R is gas constant =8.314j/k

n=mole of gas

Substituting,we have

W=p1v1-p2v2=mR(T1-T2)/n-1

M=mass of the Argon=2kg

T1=0°c=0+273=273K

T2=80+273=353k

n=polytropic index or rate of compression 1.4

Work done =2×8.314×(273-353)/1.4-1=-3,325.6joules

Negative sign means work input

Note :equation is polytropic because it is governed by the relationship pv^n=constant

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