At $N.T.P.$ one mole of diatomic gas is compressed adiabatically to half of its volume $\gamma = 1.41$. The work done on gas will be ....... $J$

Following figure shows on adiabatic cylindrical container of volume ${V_0}$ divided by an adiabatic smooth piston (area of cross-section = $A$ ) in two equal parts. An ideal gas $({C_P}/{C_V} = \gamma )$ is at pressure $P_1$ and temperature $T_1$ in left part and gas at pressure $P_2$ and temperature $T_2$ in 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)

A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then

At ${27^o}C$ a gas is suddenly compressed such that its pressure becomes $\frac{1}{8}th$ of original pressure. Temperature of the gas will be $(\gamma = 5/3)$

In an adiabatic process, the density of a diatomic gas becomes $32$ times its initial value. The final pressure of the gas is found to be $n$ times the initial pressure. The value of $n$ is

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