A thermodynamic system undergoes cyclic process $ABCDA $ as shown in figure. The work done by the system in the cycle is
${P_0}{V_0}$
$\;2{P_0}{V_0}$
$\frac{{\;{P_0}{V_0}}}{2}$
zero
Two rigid boxes containing different ideal gases are placed on a table. Box $A$ contains one mole of nitrogen at temperature $T_0$, while box $B$ contains one mole of helium at temperature $\left( {\frac{7}{3}} \right){T_0}$. The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases, $T_f$ in terms of $T_0$ is
A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac {1}{6}$ . When $T_2$ is lowered by $60\,K$ ; its efficiency increases to $\frac {1}{3}$. Then $T_1$ and $T_2$ are respectively
An ideal gas expands isothermally from a volume $V_1$ to $V_2$ and then compressed to original volume $V_1$ adiabatically. Initial pressure is $P_1$ and final pressure is $P_3$. The total work done is $W$. Then
If $\Delta Q$ and $\Delta W$ represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written as
where $\Delta U$ is the internal energy
Calculate heat given to gas during process $ABA$ in figure ...... $J$