The change in internal energy when $20 \,g$ of a gas is heated from $25^{\circ} C$ to $35^{\circ} C$ at constant volume is (Specific heat capacity of the gas at constant volume $C_{v} = 0.2 \,cal \,g^{-1} {}^{\circ} C^{-1}$): (in $\,J$)

$A$ given quantity of gas is taken from $A$ to $C$ in two ways: $a$) directly from $A \rightarrow C$ along a straight line,and $b$) in two steps,from $A \rightarrow B$ and then from $B \rightarrow C$. The work done and heat absorbed along the direct path $A \rightarrow C$ are $200 \ J$ and $280 \ J$ respectively. If the work done along the path $A \rightarrow B \rightarrow C$ is $80 \ J$,find the heat absorbed along this path.

$110\; J$ of heat is added to a gaseous system,whose internal energy change is $40\; J$,then the amount of external work done is ........ $J$.

The latent heat of vaporization of water is $2240 \, J/g$. If the work done in the process of vaporization of $1 \, g$ is $168 \, J$,then the increase in internal energy is .... $J$.

During a thermodynamic process,the pressure of a fixed mass of gas is changed in such a way that the gas releases $20 \ J$ of heat and $10 \ J$ of work is done on the gas. If the initial internal energy of the gas is $40 \ J$,then the final internal energy is ...... $J$.

$A$ perfect gas goes from state $A$ to another state $B$ by absorbing $8 \times 10^5 \ J$ of heat and doing $6.5 \times 10^5 \ J$ of external work. It is now transferred between the same two states in another process in which it absorbs $10^5 \ J$ of heat. Then in the second process: