Question

There are 1.5 Kg_m of a gas where K = 1.3 and R = 0.38 KJ/Kg_m - °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.

Answer 1

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