Question

During the thermodynamic process shown in figure for an ideal gas.

• A. $\Delta T=0$
• B. $\Delta Q=0$
• C. $W<0$
• D. $\Delta U>0$

Answer 1

The correct option is D $\Delta U>0$

Let us assume number of moles of an ideal gas is constant.


For a straight $P-V$ graph line, we can infer that $P\propto V$.
As the pressure increases, we can say from the graph that volume increases.
But, $T$ also increases because the gas chosen is ideal in nature $[\because PV\propto T]$.
Thus, we can claim that $\Delta T\ne 0$
Since are under $P-V$ graph is positive, work is considered positive [i.e $W>0$]
and as temperature is increasing, we can say that for the ideal gas, $\Delta U>0$.

Using, $\Delta Q=\Delta U+\Delta W$, we get $\Delta Q>0$
Thus, option (d) is the correct answer.