Question

.The relationship between pressure & volume in a non flow process is prescribed by the expression p=(3/V +2) ;p in a bar and volume v is in m3 . During the process 1600kj of heat is added to the gas and to the volume changes from 1.2 m3 to 4m3. Determine the change in internal energy.
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Answer 1

Change in internal energy is 953.69 kJ.

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Explanation:

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Given that

Pressure volume relationship

$P=\dfrac{3}{V}+2$

Heat added, Q= 1600 kJ

We know that, wok done is given as

$W=\int Pd V$

Now calculate the work done

$W=\int_{1.2}^{4} (\dfrac{3}{V}+2)dv$

$W=(3lnV+2V)_{1.2}^{4}\ bar.m^3$

W=646.30 kJ                         ( 1 bar =100 k Pa)

From first law of thermodynamics

Q=W + ΔU

ΔU=Change in the internal energy

W=Work transfer

Q= Heat transfer  

Therefore

1600 = 646.30 + ΔU

ΔU=953.69 kJ

Therefore the change in internal energy will be 953.69 kJ.