One mole of helium is adiabatically expanded from its initial state $({P_i}, {V_i}, {T_i})$ to its final state $({P_f}, {V_f}, {T_f})$. The decrease in the internal energy associated with this expansion is equal to

One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is $100 K$ and the universal gas constant $R = 8.0 J mol^{-1} K^{-1}$,the decrease in its internal energy,in Joule,is. . . . .

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:

$A$ cylinder with adiabatic walls is closed at both ends and is divided into two compartments by a frictionless adiabatic piston. Ideal gas is filled in both (left and right) compartments at the same $P, V, T$. Heating is started from the left side until the pressure changes to $\frac{27P}{8}$. If the initial volume of each compartment was $9 \text{ litres}$, then the final volume in the right-hand side compartment is . . . . . . litres. (for this ideal gas $\gamma = C_P/C_V = 1.5$)

$A$ reversible adiabatic path on a $P-V$ diagram for an ideal gas passes through state $A$ where $P = 0.7 \times 10^5 \, N/m^2$ and $V = 0.0049 \, m^3$. The ratio of specific heats of the gas is $1.4$. The slope of the path at $A$ is:

Is it possible to increase the temperature of a gas without adding heat to it? Explain.