c $dS = \frac{dQ}{T}$ In freezing process $dQ$ is negative so entropy decreases.
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One mole of an ideal gas at initial temperature $T$, undergoes a quasi-static process during which the volume $V$ is doubled. During the process, the internal energy $U$ obeys the equation $U=a V^3$, where $a$ is a constant. The work done during this process is
The pressure and volume of an ideal gas are related as $\mathrm{PV}^{3 / 2}=\mathrm{K}$ (Constant). The work done when the gas is taken from state $A\left(P_1, V_1, T_1\right)$ to state $\mathrm{B}\left(\mathrm{P}_2, \mathrm{~V}_2, \mathrm{~T}_2\right)$ is :
The figure, shows the graph of logarithmic reading of pressure and volume for two ideal gases $A$ and $B$ undergoing adiabatic process. From figure it can be concluded that
A heat engine operates with the cold reservoir at temperature $324 K$. The minimum temperature of the hot reservoir, if the heat engine takes $300 \; J$ heat from the hot reservoir and delivers $180 \; J$ heat to the cold reservoir per cycle, is $\dots \; K .$