One mole of an ideal monoatomic gas undergoes the following four reversible processes:
Step $1$: It is first compressed adiabatically from volume $8.0 \, m^{3}$ to $1.0 \, m^{3}$.
Step $2$: Then expanded isothermally at temperature $T_{1}$ to volume $10.0 \, m^{3}$.
Step $3$: Then expanded adiabatically to volume $80.0 \, m^{3}$.
Step $4$: Then compressed isothermally at temperature $T_{2}$ to volume $8.0 \, m^{3}$.
Then,$T_{1} / T_{2}$ is:

Two samples $A$ and $B$ of gas are initially at the same pressure and temperature. They are compressed from volume $V$ to $\frac{V}{2}$ ($A$ isothermally and $B$ adiabatically). The relation between the pressure of gas $A$ $(P_A)$ and the pressure of gas $B$ $(P_B)$ is:

$A$ heating element of resistance $r$ is fitted inside an adiabatic cylinder which carries a frictionless piston of mass $m$ and cross-sectional area $A$. The cylinder contains one mole of a diatomic gas. The temperature of the gas varies with time $t$ as $T = \alpha t + \frac{1}{2} \beta t^2$ (where $\alpha$ and $\beta$ are constants), while the pressure remains constant. The atmospheric pressure above the piston is $P_0$. Then:

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 $\left|\frac{Q_{1 \rightarrow 2}}{Q_{3 \rightarrow 4}}\right| = \frac{1}{2}$

One gram of a liquid is converted to vapour at $3 \times 10^5 \text{ Pa}$ pressure. If $10\%$ of the heat supplied is used for increasing the volume by $1600 \text{ cm}^3$ during this phase change, then the increase in internal energy in the process will be

Consider the thermodynamic cycle shown on the $PV$ diagram. The process $A \rightarrow B$ is isobaric,$B \rightarrow C$ is isochoric,and $C \rightarrow A$ is a straight-line process. The following internal energy change and heat are given: $\Delta U_{A \rightarrow B} = + 400 \text{ kJ}$ and $Q_{B \rightarrow C} = - 500 \text{ kJ}$. The heat flow in the process $Q_{C \rightarrow A}$ is ...... $\text{kJ}$.