If a gas is taken from $A$ to $C$ through $B$, the heat absorbed by the gas is $8 \, J$. Calculate the heat absorbed by the gas when it is taken from $A$ to $C$ directly.

$A$ system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in the figure. If $\Delta U_1$ and $\Delta U_2$ are the changes in internal energies in the processes $I$ and $II$ respectively,then:

An ideal gas is taken through the path $A \rightarrow B \rightarrow C \rightarrow A$ as shown in the figure. The net work done by the gas is ....... $J$.

$A$ gas takes part in two processes in which it is heated from the same initial state $1$ to the same final temperature. The processes are shown on the $P-V$ diagram by the straight lines $1-2$ and $1-3$. Points $2$ and $3$ lie on the same isothermal curve. If $Q_1$ and $Q_2$ are the heat transferred along the two processes,then:

An ideal gas undergoes a change in its state from the initial state $I$ to the final state $F$ via two possible paths as shown below. Then,

$n$ moles of a van der Waals' gas obeying the equation of state $\left(p+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$,where $a$ and $b$ are gas-dependent constants,is made to undergo a cyclic process that is depicted by a rectangle in the $p-V$ diagram as shown below. What is the heat absorbed by the gas in one cycle?