In an isobaric process,the work done by a diatomic gas is $10 \ J$. The heat given to the gas will be (in $J$):

The temperature of $3.00 \, mol$ of an ideal diatomic gas is increased by $40.0^{\circ} C$ without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of work done by the gas is $\frac{x}{10}$,then the value of $x$ (round off to the nearest integer) is ..... . (Given $R = 8.31 \, J \, mol^{-1} K^{-1}$)

Obtain the expression for work done by a gas during an isobaric process (constant pressure).

One mole of a diatomic gas is heated at constant pressure starting at $ 27^{\circ}C $. How much energy must be added to the gas as heat to double its volume?

The equation of state for a gas is given by $PV = nRT + \alpha V$,where $n$ is the number of moles and $\alpha$ is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are $T_0$ and $P_0$ respectively. The work done by the gas when its temperature doubles isobarically will be

In an isochoric process,if $t_1 = 27^{\circ}C$ and $t_2 = 127^{\circ}C$,then $\frac{P_1}{P_2}$ will be equal to [$P_1$ and $P_2$ are the pressures at $t_1^{\circ}C$ and $t_2^{\circ}C$ respectively].