Q.Boyle's law is applicable for an

The key idea here comes from the kinetic theory of gases. In an ideal gas, molecules are in constant, random motion, and their average kinetic energy is directly proportional to the absolute temperature. This is a fundamental result that does not involve pressure, volume, or density directly — those variables affect the number of collisions or the spacing of molecules, but not the average energy per molecule at a given temperature.

Why this approach works:

The kinetic theory derives the average translational kinetic energy from the root-mean-square speed of molecules, which itself comes from the ideal gas law and the pressure exerted by molecular collisions. When you work through the derivation, temperature emerges as the sole variable controlling the average kinetic energy per molecule. Pressure and volume are macroscopic properties that depend on both temperature and the number of molecules, but they do not independently change the average energy of a single molecule.

Step-by-step reasoning:

1. Recall the kinetic theory expression for pressure.

   For an ideal gas, pressure is given by $P = \frac{1}{3} \frac{N}{V} m \overline{v^2}$, where $N$ is the number of molecules, $V$ is volume, $m$ is the mass of one molecule, and $\overline{v^2}$ is the mean square speed.

2. Relate pressure to temperature via the ideal gas law.

   The ideal gas law states $PV = N k_B T$, where $k_B$ is Boltzmann’s constant. Equating this with the kinetic theory expression:

Cancelling $N/V$ from both sides gives:

3. Define the average kinetic energy. The average translational kinetic energy per molecule is $\langle K \rangle = \frac{1}{2} m \overline{v^2}$. From the equation above, $\frac{1}{2} m \overline{v^2} = \frac{3}{2} k_B T$. So:

4. Interpret the result. The formula contains only $T$ and constants. No $P$, $V$, or density appears. Therefore, the average kinetic energy of a gas molecule depends only on temperature.

Concept & Intuition

The key is that the gas is in a closed vessel — the volume cannot change. For an ideal gas at constant volume, the pressure is directly proportional to the absolute temperature (Gay-Lussac’s law). A percentage change in pressure must equal the same percentage change in absolute temperature. The trap is using Celsius degrees directly; we must convert to kelvin because the proportionality holds only for absolute temperature.

Step-by-step solution

1. State the relationship For a fixed mass of gas at constant volume:

where $T$ is absolute temperature (in kelvin).

2. Express the given changes Let the initial absolute temperature be $T_1$ (in K). The temperature increases by $2.4\,^\circ\mathrm{C}$, so

(A change of $2.4\,^\circ\mathrm{C}$ equals a change of $2.4\,\mathrm{K}$.)

The pressure increases by $0.5\%$, meaning

3. Substitute into the proportionality

Cancel $P_1$ (non-zero):

4. Solve for $T_1$ Cross-multiply:

5. Convert to Celsius

The key concept here is the relationship between the root-mean-square (rms) speed of gas molecules and temperature. From kinetic theory, the average kinetic energy of a molecule is $\frac{3}{2}kT$, and since kinetic energy is $\frac{1}{2}mv^2$, we get $v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$. This means $v_{\text{rms}} \propto \sqrt{T}$, where $T$ must be in kelvin (absolute temperature). A common mistake is to use Celsius directly — that would give a wrong answer because the zero point matters.

Let’s work through it step by step.

1. Convert temperatures to kelvin.

   The formula $v_{\text{rms}} \propto \sqrt{T}$ only holds for absolute temperature.

   $T_1 = 77^\circ\mathrm{C} = 77 + 273 = 350\,\mathrm{K}$

   $T_2 = 150.5^\circ\mathrm{C} = 150.5 + 273 = 423.5\,\mathrm{K}$

2. Write the proportionality.

   Since $v_{\text{rms}} \propto \sqrt{T}$, we have:

Here $v_1 = 50\,\mathrm{m/s}$, $T_1 = 350\,\mathrm{K}$, $T_2 = 423.5\,\mathrm{K}$.

3. Compute the ratio.

Notice that $1.21 = 1.1^2$, so $\sqrt{1.21} = 1.1$.

4. Find $v_2$.

Concept & Intuition

The key idea comes from the kinetic theory of gases. For an ideal gas, the average translational kinetic energy per molecule is given by $\frac{3}{2} k_B T$, where $k_B$ is Boltzmann’s constant and $T$ is the absolute temperature. Notice that this expression contains no dependence on molecular mass — it is purely a function of temperature. So, regardless of whether the gas is light hydrogen ($H_2$) or heavy oxygen ($O_2$), at the same temperature each molecule has the same average translational kinetic energy. The common pitfall is to confuse this with average speed or total kinetic energy per mole, which do depend on mass.

Step-by-step reasoning

1. Recall the equipartition theorem For a monatomic ideal gas, each translational degree of freedom contributes $\frac{1}{2} k_B T$ to the average kinetic energy per molecule. Since there are three translational degrees of freedom (x, y, z), the average translational kinetic energy per molecule is

2. Apply to both gases Hydrogen and oxygen are diatomic, but at ordinary temperatures their translational motion still has three degrees of freedom. Thus, for a single molecule of either gas:

3. Take the ratio Since both expressions are identical,

4. Why other options are wrong
   • Options like $1:8$ or $1:4$ would arise if one mistakenly used the ratio of average speeds (which goes as $\sqrt{1/M}$) or the ratio of total kinetic energy per mole (which is also the same because $\frac{3}{2}RT$ per mole is mass-independent).
   • The mass difference affects root-mean-square speed and diffusion rates, but not the average translational kinetic energy at a given temperature.

The key concept here is the kinetic theory of gases. For an ideal gas, the average translational kinetic energy of a single molecule is given by $\frac{3}{2} k_B T$, where $k_B$ is Boltzmann's constant and $T$ is the absolute temperature. Notice that this expression contains no term for molecular mass — it is purely a function of temperature.

This is a fundamental result: temperature is a measure of the average kinetic energy of random molecular motion. So if two different gases are at the same temperature, their molecules, on average, have the same translational kinetic energy, regardless of how heavy or light they are.

1. Write the formula for the average translational kinetic energy of a molecule:

   $\overline{K} = \frac{3}{2} k_B T$.

2. For hydrogen ($H_2$) at temperature $T$:

   $\overline{K}_H = \frac{3}{2} k_B T$.

3. For oxygen ($O_2$) at the same temperature $T$:

   $\overline{K}_O = \frac{3}{2} k_B T$.

4. The ratio is therefore:

   $\frac{\overline{K}_H}{\overline{K}_O} = \frac{\frac{3}{2} k_B T}{\frac{3}{2} k_B T} = 1$.

The rms speed is $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$. Equal rms speeds for hydrogen and oxygen require

Oxygen temperature: $T_{\text{O}} = 6495 + 273 = 6768\text{ K}$, with $M_{\text{O}} = 32$, $M_{\text{H}} = 2$.

Converting: $T_{\text{H}} = 423 - 273 = 150\,^\circ\text{C}$.

When an ideal gas undergoes an adiabatic process, it means there is no heat exchange between the gas and its surroundings. This is a crucial condition that allows us to derive a specific relationship between its state variables like pressure, volume, and temperature. We will use the first law of thermodynamics, the ideal gas law, and the definition of specific heats to establish the relationship between temperature and volume.

Here's how we derive the relation:

1. Start with the First Law of Thermodynamics for an Adiabatic Process: The first law of thermodynamics states that the heat supplied to a system ($dQ$) is used to increase its internal energy ($dU$) and do work ($dW$) on the surroundings.

For an adiabatic process, there is no heat exchange, so $dQ = 0$.

This implies that any change in internal energy must be equal to the negative of the work done by the gas.

2. Express $dU$ and $dW$ in terms of state variables: For an ideal gas, the change in internal energy ($dU$) is given by:

where $n$ is the number of moles, $C_v$ is the molar specific heat at constant volume, and $dT$ is the change in absolute temperature.
The work done by the gas ($dW$) during a small volume change $dV$ is:

where $P$ is the pressure.
Substituting these into the adiabatic first law equation:

3. Use the Ideal Gas Law to eliminate Pressure ($P$): The ideal gas law relates pressure, volume, temperature, and the number of moles:

where $R$ is the universal gas constant.
From this, we can express pressure as $P = \frac{nRT}{V}$.
Substitute this expression for $P$ into our equation:

4. Simplify and Separate Variables: We can cancel $n$ from both sides and rearrange the terms to separate temperature and volume variables:

Divide by $T$ and $C_v$:

5. Relate $R$ to $C_p$ and $C_v$ using Mayer's Relation: Mayer's relation for an ideal gas states:

where $C_p$ is the molar specific heat at constant pressure.
We are given $\gamma = \frac{C_p}{C_v}$.
Let's express $\frac{R}{C_v}$ in terms of $\gamma$:

6. Substitute and Integrate: Now substitute $\gamma - 1$ into our separated equation:

Integrate both sides:

where $\ln K'$ is the integration constant.
Rearrange the terms:

Using logarithm properties ($\alpha \ln x = \ln x^\alpha$ and $\ln x + \ln y = \ln (xy)$):

Exponentiating both sides gives:

where $K$ is a constant.

Comparing this derived relation with the given options:

(A) $TV = \text{Constant}$

(B) $TV^{\gamma} = \text{Constant}$

(C) $TV^{\gamma-1} = \text{Constant}$

(D) $\frac{T}{V^{\gamma}} = \text{Constant}$

The derived relation matches option (C).

The problem hinges on two fundamental facts about an ideal gas. First, the difference between the molar specific heats at constant pressure and constant volume is always equal to the universal gas constant: $C_p - C_v = R$. Second, the ratio $\gamma = C_p/C_v$ is a constant for a given gas (e.g., about 1.67 for monatomic, 1.4 for diatomic). These two equations together let us express $C_v$ purely in terms of $\gamma$ and $R$.

Why does $C_p - C_v = R$ hold? At constant volume, all heat added goes into increasing internal energy (no work done). At constant pressure, the gas expands as it heats, doing work on the surroundings, so more heat is needed to raise the temperature by the same amount — that extra heat exactly equals $R$ per mole per kelvin. This is a cornerstone result from kinetic theory and thermodynamics.

Now, let’s solve step by step.

1. Write the two relations. We have:

and

2. Substitute (2) into (1). Replace $C_p$ in equation (1) with $\gamma C_v$:

3. Factor out $C_v$.

4. Solve for $C_v$. Divide both sides by $(\gamma - 1)$:

The root mean square speed of a gas is a direct measure of the average kinetic energy of its molecules. For an ideal gas, the kinetic theory tells us that the average translational kinetic energy per molecule is $\frac{3}{2}kT$, which leads to the famous relation:

Here $R$ is the universal gas constant, $T$ is the absolute temperature (in Kelvin), and $M$ is the molar mass (in kg/mol). The key insight: at a given temperature, lighter molecules move faster. So to make a heavier gas (Neon) have the same rms speed as a lighter gas (Helium), we must heat the heavier gas to a higher temperature.

Let’s work through it step by step.

1. Convert the given temperature to Kelvin Helium is at $-33~^\circ \mathrm{C}$.

2. Write the equality condition We want $v_{\text{rms, Ne}} = v_{\text{rms, He}}$. Using the formula:

3. Cancel common factors The $3R$ cancels from both sides. Squaring both sides gives:

4. Solve for $T_{\text{Ne}}$

Compute: $\frac{20.2}{4.0} = 5.05$, so

5. Check the options The result matches option (C) exactly.

The root-mean-square (RMS) velocity of gas molecules is a measure of their average speed. It's not simply the arithmetic mean because molecules move in random directions and at varying speeds. The RMS velocity gives a more representative average, especially when considering kinetic energy.

The kinetic theory of gases establishes a direct relationship between the average kinetic energy of gas molecules and the absolute temperature of the gas. Since kinetic energy depends on mass and velocity, this means temperature directly influences molecular speed. Specifically, for an ideal gas, the RMS velocity is given by:

This formula tells us that for a given gas, $v_{rms}$ increases with the square root of temperature. Conversely, for a given temperature, lighter gases (smaller $M$) have higher $v_{rms}$ than heavier gases (larger $M$).

In this problem, we are asked to find the temperature at which oxygen molecules have the same RMS velocity as hydrogen molecules at a specific temperature. This means we will set the $v_{rms}$ expressions for both gases equal to each other and solve for the unknown temperature.

Here's how to solve the problem step-by-step:

1. Identify the given information and the goal:

   • For hydrogen ($H_2$): Temperature $T_H = 20 \text{ K}$.
   • For oxygen ($O_2$): We need to find the temperature $T_O$ such that $v_{rms,O} = v_{rms,H}$.

2. Determine the molar masses of hydrogen and oxygen:

   • Hydrogen is a diatomic molecule, $H_2$. The atomic mass of hydrogen is approximately $1 \text{ g/mol}$. So, the molar mass of $H_2$ is $M_H = 2 \times 1 \text{ g/mol} = 2 \text{ g/mol}$. Converting to $\text{kg/mol}$: $M_H = 2 \times 10^{-3} \text{ kg/mol}$.
   • Oxygen is a diatomic molecule, $O_2$. The atomic mass of oxygen is approximately $16 \text{ g/mol}$. So, the molar mass of $O_2$ is $M_O = 2 \times 16 \text{ g/mol} = 32 \text{ g/mol}$. Converting to $\text{kg/mol}$: $M_O = 32 \times 10^{-3} \text{ kg/mol}$.

3. Set the RMS velocities equal:

   We are given that $v_{rms,O} = v_{rms,H}$. Using the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$:

4. Solve for $T_O$: Square both sides of the equation to remove the square roots:

Notice that $3R$ appears on both sides, so we can cancel it out:

Now, rearrange the equation to solve for $T_O$:

5. Substitute the values and calculate: Substitute the known values for $T_H$, $M_O$, and $M_H$:

The units $\text{kg/mol}$ cancel out, as do the factors of $10^{-3}$:

The temperature at which an oxygen molecule has the same RMS velocity as a hydrogen molecule at 20 K is 320 K. This corresponds to option (B).

Setup. A sudden burst is an adiabatic (fast, no heat exchange) expansion. Initial state: $P_1 = 2$ atm, $T_1 = 27^\circ\text{C} = 300$ K. Final: $P_2 = 1$ atm, with $\gamma = C_p/C_v = 1.5$.

Adiabatic pressure–temperature relation.

The expansion drops the temperature by the factor the problem supplies, $2^{2/3} = 1.2$:

Back to Celsius.

The air cools on expansion, so the negative value is physically expected.