The change in the internal energy of a mass of gas,when the volume changes from $V$ to $2V$ at constant pressure $P$,is (where $\gamma$ is the ratio of specific heat of gas at constant pressure to specific heat at constant volume):

If $\Delta Q$ and $\Delta W$ represent the heat supplied to the system and the work done on the system respectively,then the first law of thermodynamics can be written as,where $\Delta U$ is the change in internal energy.

$A$ gas absorbs $18 \ J$ of heat and work done on the gas is $12 \ J$. Then the change in internal energy of the gas is: (in $J$)

Write the sign convention for internal energy in thermodynamics.

$A$ diatomic gas has an initial internal energy of $80 cal$. $A$ work of $18 cal$ is done on the gas and the gas releases heat energy of $42 J$. The final internal energy of the gas is

Consider $1 \,kg$ of liquid water undergoing a change in phase to water vapour at $100^{\circ} C$. At $100^{\circ} C$,the vapour pressure is $1.01 \times 10^5 \,N m^{-2}$ and the latent heat of vaporization is $22.6 \times 10^5 \,J kg^{-1}$. The density of liquid water is $10^3 \,kg m^{-3}$ and that of vapour is $\frac{1}{1.8} \,kg m^{-3}$. The change in internal energy in this phase change is nearly ............ $J kg^{-1}$.