In a thermodynamic process, helium gas obeys the law $TP^{-2/5} = \text{constant}$. The heat given to the gas when the temperature of $2$ moles of the gas is raised from $T$ to $4T$ ($R$ is the universal gas constant) is:

One mole of an ideal gas at an initial temperature of $T$ $K$ does $6R$ of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5/3$,the final temperature of the gas will be $(R=8.31 \ J \ mole^{-1} \ K^{-1})$.

$A$ 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:

The pressure and density of a diatomic gas $(\gamma = 7/5)$ change adiabatically from $(P, d)$ to $(P', d')$. If $\frac{d'}{d} = 32$,then $\frac{P'}{P}$ is equal to:

During an adiabatic expansion of $2\, moles$ of a gas,the change in internal energy was found to be $-50\, J$. The work done during the process is ...... $J$.

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. . . . .