$2$ moles of an ideal monoatomic gas is carried from a state $(p_{0}, V_{0})$ to state $(2 p_{0}, 2 V_{0})$ along a straight line path in a $p-V$ diagram. The amount of heat absorbed by the gas in the process is given by

What is an isothermal process, adiabatic process, and isobaric process? Write the first law of thermodynamics for an ideal gas.

In the $P-V$ diagram shown,the gas does $5 \, J$ of work in the isothermal process $ab$ and $4 \, J$ of work in the adiabatic process $bc$. What will be the change in internal energy of the gas in the straight path $c$ to $a$ (in $, J$)?

An ideal monoatomic gas is carried along the cycle $ABCDA$ as shown in the figure. The total heat absorbed during this process is (in $p_0 V_0$)

Carbon monoxide is taken around the closed cycle $abc$. The process $bc$ is isothermal as shown in the figure. When the temperature of the gas is increased from $300 \ K$ to $1000 \ K$,it absorbs $7000 \ J$ of heat in going from $a$ to $b$. The heat released by the gas during the process $ca$ is ...... $J$.

Consider one mole of helium gas enclosed in a container at initial pressure $P_1$ and volume $V_1$. It expands isothermally to volume $4 V_1$. After this,the gas expands adiabatically and its volume becomes $32 V_1$. The work done by the gas during isothermal and adiabatic expansion processes are $W_{\text{iso}}$ and $W_{\text{adia}}$,respectively. If the ratio $\frac{W_{\text{iso}}}{W_{\text{adia}}} = f \ln 2$,then $f$ is: