Solution;
Give;
p1=1.02bar
T1=22°c=295K
V1=0.015m3
p2=6.8bar
Law of compression;
pVγ=C
(i)Final temperature;
Using the relation;
T1T2=(p1p2)γγ−1
γ for air is 1.4;
295T2=(1.026.8)1.41.4−1
T2=295(1.026.9)1.40.4 =507.24K
Final temperature is;
507.24-273=234.24°c
(ii)Final volume;
Using the relation;
p1V1γ=p2V2γ
Resolve as;
V1V2=(p2p1)γ1
V2=0.015×(6.81.02)1.41
V2=0.00387m3
(iii)Work done;
Using;
W=γ−1mR(T2−T1)
m is the mass found by the following relation;
pV=mRT
m=RT1p1V1 =287×2951.02×105×0.015
m=0.01807kg
Hence;
W=1.4−10.01807×287×(295−507.24)
W=−2751J
Negative sign means work is done on the air.
Comments