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# Answer to Question #165480 in Mechanical Engineering for Rodel Basco

Question #165480

There are 1.5 Kgm of a gas where K = 1.3 and R = 0.38 KJ/Kgm - °R that undergo an isochoric process from p1 = 0.552 MPa, t1 = 58.5°C top2 = 1.66 MPa. During the process, there added 100 KJ of heat. Compute the heat transferred, change of internal energy and the change of entropy.

1
2021-02-26T05:41:22-0500

mass of gas=1.5 Kg,K=1.3, and R=0.38 KJ/Kgm - °R and

P1 = 0.552 MPa, T1 = 58.5°C ,P2 = 1.66 MPa., Heat added to system = 100 kJ

As the process is isochoric , we know that for isochoric process work done is zero.

and heat transfer is given as , "\\Delta Q= 100 kJ"

As per first law of thermodynamics

"\\Delta Q= \\Delta U+ \\Delta W"

100 = "\\Delta U +0"

So, internal energy = 100 kJ

"\\Delta S= \\frac{ \\Delta Q}{T}= \\frac{ 100 \\times 1000}{ 331.5 }"

"\\Delta S= 301.659 J\/K" is the entropy of system

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