An ideal gas is found to obey $p V^{3/2} = \text{constant}$ during an adiabatic process. If such a gas initially at a temperature $T$ is adiabatically compressed to half of its initial volume, then its final temperature is

The relation between volume $(V)$ and absolute temperature $(T)$ of a gas in an adiabatic process is

You feel refreshed by taking a shower in summer but not in winter. Why?

An ideal gas has a specific heat capacity at constant pressure of $\frac{11}{10} R$. If one mole of this ideal gas at $125^{\circ} C$ does $83 \,J$ of work adiabatically, then the final temperature of the gas would be (Universal gas constant, $R=8.3 \,J \,K^{-1} \,mol^{-1}$ ). (in $^{\circ} C$)

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..... $K$ $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

The pressure and density of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$ changes adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d}=32$,then $\frac{P^{\prime}}{P}$ is: