Gas Laws
By Gas laws describe the behavior of gases under various conditions, including changes in temperature, pressure, and volume. These laws are based on the concepts of thermodynamics and the equation of state, which relates the state variables of a gas. The common gas laws are:
Boyle's Law
Boyle's law describes the relationship between the volume and pressure of a gas at constant temperature. It states that the product of the pressure and volume of a gas is constant when the temperature is kept constant i.e., \begin{align} P_1 V_1 = P_2 V_2 \end{align} where $P_1$ and $V_1$ are the initial pressure and volume of the gas, and $P_2$ and $V_2$ are the final pressure and volume, respectively.
Charle's Law
Charles's law describes the relationship between the volume and temperature of a gas at constant pressure. It states that the volume of a gas is directly proportional to its temperature when the pressure is kept constant i.e., \begin{align} \frac{V_1}{T_1} = \frac{V_2}{T_2} \end{align} where $V_1$ and $T_1$ are the initial volume and temperature of the gas, and $V_2$ and $T_2$ are the final volume and temperature, respectively.
Gay-Lussac's Law
Gay-Lussac's Law describes the relationship between the pressure and temperature of a gas at constant volume. It states that the pressure of a gas is directly proportional to its temperature when the volume is kept constant i.e., \begin{align} \frac{P_1}{T_1} = \frac{P_2}{T_2} \end{align} where $P_1$ and $T_1$ are the initial pressure and temperature of the gas, and $P_2$ and $T_2$ are the final pressure and temperature, respectively.
Ideal Gas Law
The ideal gas law, which combines all three gas laws, can be used to describe the behavior of ideal gases. It relates the pressure, volume, temperature, and number of moles of a gas and is given by the equation \begin{align} PV = nRT \end{align} where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles of gas, $R$ is the gas constant, and $T$ is the temperature of the gas in Kelvin.
Problems from IIT JEE
Problem (IIT JEE 1978): A column of mercury of length 10 cm is contained in the middle of a horizontal tube of length 1 m which is closed at both the ends. The two equal lengths contain air at standard atmospheric pressure of 0.76 m of mercury. The tube is now turned to vertical position. By what distance will the column of mercury be displaced? Assume temperature to be constant.
Solution: Let A and B be the portions of the tube on the left and the right sides of the mercury column. In horizontal condition, both A and B have equal lengths $l_0=45$ cm and equal pressures $p_0=0.76$ m of Hg.
Let mercury column is dips down by a distance $x$ when tube is turned to the vertical position. Lengths of A and B change to $l_A=(45-x)$ cm and $l_B=(45+x)$ cm. Let the pressures in A and B change to $p_A$ and $p_B$. As temperature is constant, apply Boyle's law, $p_iV_i=p_fV_f$, on A and B to get \begin{align} &p_0(45A)=p_A(45-x)A,\\ &p_0(45A)=p_B(45+x)A, \end{align} where $A$ is the cross-sectional area of the tube. In the vertical position, the pressure at the lower face of mercury ($p_A$) is equal to the sum of the pressure at the upper face ($p_B$) and the hydrostatic pressure due to mercury column of height 10 cm, i.e., \begin{align} p_A=p_B+10\rho g. \end{align} Eliminate $p_A$ and $p_B$ from above equations to get \begin{align} x &=\frac{45 (10 \rho g)}{2p_0} \\ &=\frac{45 (10 \rho g)}{2 (76 \rho g)} \\ &=2.96\;\mathrm{cm}.\nonumber \end{align}
