Heat and Thermodynamics

From $P$ - $V$ diagram, Process 1: $P=$ constant, Volume $(V)$ increases and temperature $(T)$ also increases. Work done $W=\int P dV=$ positive, $\Delta U=\frac{f}{2} n R \Delta T=$ positive Therefore heat is absorbed by the gas. $\because \Delta Q=\Delta U+W$ Process -2: Volume $(V)=$ Constant, Pressure $(P)$ decreases. $\therefore$ Temperature $(T)$ decreases. $\because P \propto T$ $\therefore \;\; W=\int P dV=0$ $\Delta T$ is negative and $\Delta U=\frac{f}{2} n R \Delta T$ $\Delta Q=\Delta U+W=$ Negative Heat is rejected by the gas. Process 3: $P=$ Constant, Volume $(V)$ decreases. $\therefore$ Temperature $(T)$ also decreases. $W=P \Delta V=$ Negative $\Delta U=\frac{f}{2} n R \Delta T=$ negative $\therefore \Delta Q=W+\Delta U=$ negative $\therefore$ Heat is rejected by the gas. Process 4: $V=$ constant, Pressure $(P)$ increases. $\therefore \;\; W=\int P dV=0$ $P V=n R T \Rightarrow$ Temperature increases $\therefore \;\; \Delta U=\frac{f}{2} n R \Delta T$ is positive $\therefore \Delta Q=\Delta U+W=$ positive Therefore heat is absorbed by the gas.