Question

A stationary mass of gas is compressed without friction from an initial state of 0.3 ${{m}^{3}}$ and 0.15 ${{m}^{3}}$ and 0.105M Pa. The pressure remains constant. During the process there is a transfer of 37.6 kJ of heat from the system. During the process the amount of internal energy changes is________.
A. -28.85 kJ
B. -21.85 kJ
C. -21.85 MJ
D. -53.35kJ

Answer 1

There is a relationship between change in internal energy and total amount of work done and it is as follows.
$q=\Delta U+W$
Here q = heat transfer
$\Delta U$ = change in internal energy
W =Work done

Complete step-by-step answer: In the question it is given that to calculate the change in internal energy of the system.
- In the question it is given that initial volume ${{V}_{1}}=0.3{{m}^{3}}$ and the final volume of the gas ${{V}_{2}}=0.15{{m}^{3}}$ .
- Therefore the change in volume will be ${{V}_{2}}-{{V}_{1}}=0.15-0.3=-0.15$ .
- In the question it is given that the pressure is constant = 0.015M Pa = $0.105\times {{10}^{3}}KPa$
- Again it is given that the total amount of heat transfer q = -37.6 kJ.
- As per the first law of thermodynamics there is a relationship between change in internal energy and total amount of work done and it as follows.
$q=\Delta U+W$
- The total work done w = $P\Delta V=0.105\times 1000\times -0.15=-15.75kJ$
- Now we can calculate the change in internal energy and it is as follows.

& \Delta U=q-W \\\ & \Delta U=-37.6-(-15.75) \\\ & \Delta U=-21.85kJ \\\ \end{aligned}$$ **Therefore the correct option is B.** **Note:** Initially we have to calculate the change in volume of the gas, next we have to calculate the total work done and at last we have to calculate the change in internal energy by using the given data accurately.