The temperature-entropy $(T-S)$ diagram of a reversible engine cycle is given in the figure. Its efficiency is

One mole of an ideal monoatomic gas is heated at a constant pressure of one atmosphere from $0^{\circ}C$ to $100^{\circ}C$. Then the change in the internal energy is

In the $p-V$ diagram below,the dashed curved line is an adiabat. For a process that is described by a straight line joining two points $X$ and $Y$ on the adiabat (solid line in the diagram),heat is (Hint: consider the variation in temperature from $X$ to $Y$ along the straight line)

$A$ monoatomic gas of $n$-moles is heated from temperature $T_1$ to $T_2$ under two different conditions: $(i)$ at constant volume and (ii) at constant pressure. The change in internal energy of the gas is

An ideal gas is heated at constant pressure. If $Q$ is the amount of heat absorbed,then the ratio of the work done to the heat absorbed for increasing the internal energy is:

One mole of a monoatomic ideal gas goes through a thermodynamic cycle,as shown in the volume versus temperature $(V-T)$ diagram. The correct statement$(s)$ is/are :
[$R$ is the gas constant]
$(1)$ Work done in this thermodynamic cycle $(1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1)$ is $|W| = \frac{1}{2} RT_0$
$(2)$ The ratio of heat transfer during processes $1 \rightarrow 2$ and $2 \rightarrow 3$ is $\left|\frac{Q_{1 \rightarrow 2}}{Q_{2 \rightarrow 3}}\right| = \frac{5}{3}$
$(3)$ The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
$(4)$ The ratio of heat transfer during processes $1 \rightarrow 2$ and $3 \rightarrow 4$ is $\left|\frac{Q_{1 \rightarrow 2}}{Q_{3 \rightarrow 4}}\right| = \frac{1}{2}$