In an isothermal change, an ideal gas obeys:

An ideal gas is trapped inside a test tube of cross-sectional area $20 \times 10^{-6} \, m^2$ as shown in the figure. The gas occupies a height $L_1$ at the bottom of the tube and is separated from air at atmospheric pressure by a mercury column of mass $0.002 \, kg$. If the tube is quickly turned isothermally,upside down,the gas now occupies height $L_2$ in the tube. Find the ratio $L_2/L_1$. [Take atmospheric pressure $P_0 = 10^5 \, Nm^{-2}$ and $g = 10 \, ms^{-2}$]

During an isothermal expansion,a confined ideal gas does $-150 \, J$ of work against its surroundings. This implies that

An ideal gas is subjected to an isothermal expansion such that its volume changes from $V_i$ to $V_f$ and pressure from $P_i$ to $P_f$. The work done on the gas is:

The temperature of an ideal gas is kept constant as it expands. The gas does external work. During this process,the internal energy of the gas

One mole of an ideal gas at $27^{\circ} {C}$ is taken from $A$ to $B$ as shown in the given $PV$ indicator diagram. The work done by the system will be $...... \times 10^{-1} \, {J}$.
[Given: $R = 8.3 \, {J} / \text{mole} \cdot {K}, \ln 2 = 0.6931$] (Round off to the nearest integer)