Question

A stationary mass of gas is compressed without friction from an initial stage of 0.3 m 3 and 0.105 mpa to a final state of 0.15 m3 , the pressure remaining constant. There is a transfer of 37.6 kj of heat from the gas during the process. How much does the internal energy of the gas change?

Answer 1

Answer:

Explanation:

Solution :

ΔW=−105×(0.15−0.3)KJ=+15.75KJ

Δq=−37.6KJ

∴ΔE=ΔW+Δq=−37.6+15.75=−21.85KJ

Answer 2

The internal energy of the gas is -21.85 kJ .

We know , work done in constant pressure is given by :

$W=P\Delta V\\W=P(V_f-V_i)$

Putting value of final ,initial volume and pressure in above equation we get :

$W=-0.105(0.15-0.3)=15.75\ kJ$

It is also given that heat transfer is $H=-37.6\ kJ$ .

We know , by first law of thermodynamics :

$\Delta U=W+q$     ........( 2 )

Here , $\Delta U$ is change in internal energy .

Putting all values in equation 2 , we get :

$\Delta U=15.75+(-37.6)\\\Delta U=-21.85\ kJ$

Hence , this is the required solution .

Learn More :

Thermodynamics

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