The change in the entropy of $1$ mole of an ideal gas which undergoes an isothermal process from an initial state $(P_1, V_1, T)$ to the final state $(P_2, V_2, T)$ is equal to

If an ideal gas is compressed isothermally,then:

The thermodynamic process,in which the internal energy of the system remains constant is ...........

$A$ certain gas is isothermally compressed to $(1/3)^{rd}$ of its initial volume $(V_0 = 3 \ \text{L})$ by applying required pressure. If the bulk modulus of the gas is $3 \times 10^5 \ \text{N/m}^2$, the magnitude of work done on the gas is . . . . . . $J$.

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 $g = 10 \, ms^{-2}$]

Heat is given to an ideal gas in an isothermal process.
$A.$ Internal energy of the gas will decrease.
$B.$ Internal energy of the gas will increase.
$C.$ Internal energy of the gas will not change.
$D.$ The gas will do positive work.
$E.$ The gas will do negative work.
Choose the correct answer from the options given below: