Question #235097

Air at 1.02 bar, 22°C, initially occupying a cylinder volume of 0.015 m3, is

compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate :

(i) The final temperature ;

(ii) The final volume ;

(iii) The work done.


1
Expert's answer
2021-09-10T08:54:41-0400

Solution;

Give;

p1=1.02barp_1=1.02 bar

T1=22°c=295KT_1=22°c=295K

V1=0.015m3V_1=0.015m^3

p2=6.8barp_2=6.8bar

Law of compression;

pVγ=CpV^\gamma=C

(i)Final temperature;

Using the relation;

T2T1=(p2p1)γ1γ\frac{T_2}{T_1}=(\frac{p_2}{p_1})^{\frac{\gamma-1}{\gamma}}

γ\gamma for air is 1.4;

T2295=(6.81.02)1.411.4\frac{T_2}{295}=(\frac{6.8}{1.02})^{\frac{1.4-1}{1.4}}

T2=295(6.91.02)0.41.4T_2=295(\frac{6.9}{1.02})^{\frac{0.4}{1.4}} =507.24K

Final temperature is;

507.24-273=234.24°c

(ii)Final volume;

Using the relation;

p1V1γ=p2V2γp_1V_1^\gamma=p_2V_2^\gamma

Resolve as;

V2V1=(p1p2)1γ\frac{V_2}{V_1}=(\frac{p_1}{p_2})^\frac{1}{\gamma}

V2=0.015×(1.026.8)11.4V_2=0.015×(\frac{1.02}{6.8})^\frac{1}{1.4}

V2=0.00387m3V_2=0.00387m^3

(iii)Work done;

Using;

W=mR(T2T1)γ1W=\frac{mR(T_2-T_1)}{\gamma-1}

m is the mass found by the following relation;

pV=mRTpV=mRT

m=p1V1RT1m=\frac{p_1V_1}{RT_1} =1.02×105×0.015287×295\frac{1.02×10^5×0.015}{287×295}

m=0.01807kgm=0.01807kg

Hence;

W=0.01807×287×(295507.24)1.41W=\frac{0.01807×287×(295-507.24)}{1.4-1}

W=2751JW=-2751J

Negative sign means work is done on the air.






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