Question

A thermodynamic process is shown in the given figure. The pressures and volume corresponding to some points in the figure are:
$P_A=3\times {10}^4\text{Pa}$, $P_B=8\times {10}^4\text{Pa}$
$V_A=2\times {10}^{-3}{\text{m}}^3$, $V_D=5\times {10}^{-3}{\text{m}}^3$


In the process AB, 600 J of heat is added to the system and in BC, 200 J of heat is added to the system. The change in internal energy of the system in the process AC would be:

• A. 560 J
• B. 800 J
• C. 600 J
• D. 640 J

Answer 1

The correct option is A 560 J

No work is done along the path AB because this process is isochoric (for isochoric process $\Delta V=0$)
$\therefore$ Work done $=P\Delta V=0$
Thus, the work done (W) $=-P_B(V_D-V_A)$
$=-8\times {10}^4(5\times {10}^{-3}-2\times {10}^{-3})$
$=-8\times {10}^4\times 3\times {10}^{-3}\text{J}$
$=-240\text{J}$
$\to$ The energy absorbed by the system
$=(dq{)}_{AB}+(dq{)}_{BC}=600+200=800\text{J}$
$\to$ The change in internal energy, $dE=dq+W$
$dE=800-240=560\text{J}$