The work of $146\ kJ$ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by $7^o  C$. The gas is $(R=8.3\ J\ mol^{-1} K^{-1})$

Consider one mole of helium gas enclosed in a container at initial pressure $P_1$ and volume $V_1$. It expands isothermally to volume $4 V_1$. After this, the gas expands adiabatically and its volume becomes $32 V_1$. The work done by the gas during isothermal and adiabatic expansion processes are $W_{\text {iso }}$ and $W_{\text {adia, }}$ respectively. If the ratio $\frac{W_{\text {iso }}}{W_{\text {adia }}}=f \ln 2$, then $f$ is. . . . . . . .

An ideal gas goes through a reversible cycle $a\to b\to c\to d$ has the $V – T$ diagram shown below. Process $d\to a$ and $b\to c$ are adiabatic….  The corresponding $P – V$ diagram for the process is (all figures are schematic and not drawn to scale)

$Assertion :$ Adiabatic expansion is always accompanied by fall in temperature.
$Reason :$ In adiabatic process, volume is inversely proportional to temperature.

An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes $32$ times of its initial value. If the final pressure of gas is $128$ atmosphers, the value of $\gamma$ the gas is