$A$ monoatomic gas $\left( \gamma = \frac{5}{3} \right)$ is suddenly compressed to $\frac{1}{8}$ of its original volume. What will be the final pressure of the gas in terms of its initial pressure?

The initial pressure and volume of a given mass of an ideal gas (with $\frac{C_{p}}{C_{V}}=\gamma$), taken in a cylinder fitted with a piston, are $p_{0}$ and $V_{0}$ respectively. At this stage, the gas has the same temperature as that of the surrounding medium, which is $T_{0}$. It is adiabatically compressed to a volume equal to $\frac{V_{0}}{2}$. Subsequently, the gas is allowed to come to thermal equilibrium with the surroundings. What is the heat released to the surrounding?

During an adiabatic process,the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of $\frac{C_P}{C_V}$ for the gas is

The volume of an ideal gas increases $8$ times and the temperature becomes $1/4$ of the initial temperature during a reversible adiabatic change. If there is no exchange of heat in this process $( \Delta Q=0 )$, identify the gas from the following options:

$A$ monoatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume is $TV^{x} = \text{constant}$, then $x$ is

The following figure shows an adiabatic cylindrical container of volume $V_0$ divided by an adiabatic smooth piston (area of cross-section = $A$) into two equal parts. An ideal gas $(C_P/C_V = \gamma)$ is at pressure $P_1$ and temperature $T_1$ in the left part,and a gas at pressure $P_2$ and temperature $T_2$ is in the right part. The piston is slowly displaced and released at a position where it can stay in equilibrium. The final pressure of the two parts will be (Suppose $x$ = displacement of the piston):