$A$ gas at $37^{\circ} C$ is compressed adiabatically to half of its volume. What is the final temperature of the gas (in $^{\circ} C$)? (Ratio of specific heat capacities of the gas is $1.5$)

$A$ work of $166.28 \ J$ is done to adiabatically compress one mole of a gas. If the increase in the temperature of the gas is $8^{\circ} C$,the gas is $\left(R=8.314 \ J \ mol^{-1} \ K^{-1}\right)$

$A$ gas at normal temperature is suddenly compressed to one-fourth of its original volume. If $\gamma = 1.5$, then the increase in the temperature of the gas in Kelvin is ($\gamma$ is the ratio of specific heats).

During an adiabatic compression,$830 \ J$ of work is done on $2 \ moles$ of a diatomic ideal gas to reduce its volume by $50\%$. The change in its temperature is nearly..... $K$ $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

For an adiabatic process with $\gamma = 2.5$,if the volume of a gas is reduced to $1/8$ of its initial volume,what will be the new pressure $P'$ in terms of the initial pressure $P$?

An ideal gas undergoes an adiabatic process obeying the relation $PV^{4/3} = \text{constant}$. If its initial temperature is $300 \ K$ and its pressure is increased up to four times its initial value, then the final temperature is (in Kelvin):