Question #100063
Mass of gas at an initial pressure of 28 and with an internal energy of 1500 kilo is contained in a well insulated cylinder of volume 006 metre cube. the gas is allowed to expand behind the Piston until its internal energy is 1400 kilo the law of expansion is PV square is equal to constant calculate the work done the final volume, the final pressure
1
Expert's answer
2019-12-10T09:51:59-0500

1)


W=U2U1=14001500=100 kJW=U_2-U_1=1400-1500=-100\ kJ

2)


pV2=kpV^2=k

W=12pdV=12kV2dV=k(1Vf1Vi)W=\int_1^2pdV=\int_1^2\frac{k}{V^2}dV=k\left(\frac{1}{V_f}-\frac{1}{V_i}\right)

100000=(2800000)(0.06)2(1Vf10.06)-100000=(2800000)(0.06)^2\left(\frac{1}{V_f}-\frac{1}{0.06}\right)

Vf=0.148 m3V_f=0.148\ m^3

3)


(2800000)(0.06)2=(pf)(0.148)2(2800000)(0.06)^2=(p_f)(0.148)^2

pf=459000 Pa=4.59 barp_f=459000\ Pa=4.59\ bar


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Comments

Ayodeji
04.05.21, 22:27

Thankssssssss

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