Monoatomic, diatomic and triatomic gases whose initial volume and pressure are same, are compressed till their volume becomes half the initial volume.

One mole of the ideal gas goes through the process $p=p_0\left[1-\alpha\left(\frac{V}{V_0}\right)^3\right]$, where $p$ and $V$ are pressure and volume, $p_0, V_0$ and $\alpha$ are constants. If the maximum attainable temperature of the gas is $\left(\frac{3}{4}\right) \frac{p_0 V_0}{R}$, then the value of $\alpha$ is

As shown schematically in the figure,two vessels contain water solutions (at temperature $T$) of potassium permanganate $(KMnO_4)$ of different concentrations $n_1$ and $n_2$ $(n_1 > n_2)$ molecules per unit volume with $\Delta n = (n_1 - n_2) \ll n_1$. When they are connected by a tube of small length $\ell$ and cross-sectional area $S$,$KMnO_4$ starts to diffuse from the left to the right vessel through the tube. Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion. The speed $v$ of the molecules is limited by the viscous force $-\beta v$ on each molecule,where $\beta$ is a constant. Neglecting all terms of the order $(\Delta n)^2$,which of the following is/are correct? ($k_B$ is the Boltzmann constant)
$(A)$ the force causing the molecules to move across the tube is $\Delta n k_B T S$
$(B)$ force balance implies $n_1 \beta v \ell = \Delta n k_B T$
$(C)$ total number of molecules going across the tube per sec is $\left(\frac{\Delta n}{\ell}\right)\left(\frac{k_B T}{\beta}\right) S$
$(D)$ rate of molecules getting transferred through the tube does not change with time

$125 \, ml$ of gas $A$ at $0.60 \, atm$ and $150 \, ml$ of gas $B$ at $0.80 \, atm$ pressure at the same temperature are filled in a vessel of $1 \, L$ volume. What will be the total pressure of the mixture at the same temperature in $atm$?

The difference between volume and pressure coefficients of an ideal gas is

If $\Delta Q$ is the amount of heat supplied to $n$ moles of a rigid diatomic gas at constant pressure, $\Delta U$ is the change in internal energy, and $\Delta W$ is the work done, then $\Delta W : \Delta U : \Delta Q$ is