FBISE, NCP & NBF-aligned resources for Classes 9–12
Browse by Class
Browse by Type
Physics Chapter 8 : Topic 4
Boltzmann Constant
1 / 29
The Boltzmann constant $\mathbf{(k)}$ is related to the universal gas constant $\mathbf{(R)}$ by:
Reasoning: The Boltzmann constant is obtained by dividing the universal gas constant by Avogadro’s number.
2 / 29
In the expression $\mathbf{k}\mathbf{=}$ $\frac{\mathbf{R}}{\mathbf{N}_{\mathbf{A}}}$, $\mathbf{N}_{\mathbf{A}}\mathbf{\ }$represents:
Reasoning: $\mathbf{N}_{\mathbf{A}}$ is Avogadro’s number, approximately $\mathbf{6.02 \times}\mathbf{10}^{\mathbf{23}}$.
3 / 29
The value of the Boltzmann constant is approximately:
Reasoning: The accepted value of Boltzmann constant is $\mathbf{1.38 \times}\mathbf{10}^{\mathbf{- 23}}\mathbf{\ }\mathbf{J\ K}^{\mathbf{- 1}}$.
4 / 29
The SI unit of the Boltzmann constant is:
Reasoning: Boltzmann constant relates energy per particle to temperature, so its unit is $\mathbf{J\ K}^{\mathbf{- 1}}$.
5 / 29
The universal gas constant $\mathbf{(R)}$ has SI unit:
Reasoning: $\mathbf{(R)}$ relates energy to moles and temperature.
6 / 29
The difference between $\mathbf{(R)}$ and $\mathbf{(k)}$ is that:
Reasoning: $\mathbf{(R)}$ is for one mole, while $\mathbf{(k)}$ is for one particle (molecule).
7 / 29
Which quantity connects macroscopic gas behavior to molecular behavior?
Reasoning: Boltzmann constant links molecular-scale energy with temperature.
8 / 29
If $\mathbf{R = 8.31\ J\ }\mathbf{mol}^{\mathbf{- 1}}\mathbf{K}^{\mathbf{- 1}}$ and $\mathbf{N}_{\mathbf{A}}\mathbf{= 6.02 \times}\mathbf{10}^{\mathbf{23}}$, then $\mathbf{k}$ is approximately:
Reasoning: $\mathbf{k =}$ $\frac{\mathbf{8.31}}{\mathbf{6.02\ \times \ }\mathbf{10}^{\mathbf{23}}}\mathbf{\approx 1.38 \times}\mathbf{10}^{\mathbf{- 23}}$
9 / 29
Which equation uses the Boltzmann constant directly?
Reasoning: The molecular form of the ideal gas equation contains $\mathbf{(k)}$.
10 / 29
Which relation is dimensionally correct?
Reasoning: Rearranging $\mathbf{k =}$ $\frac{\mathbf{R}}{\mathbf{N}_{\mathbf{A}}}$ gives $\mathbf{R =}\mathbf{kN}_{\mathbf{A}}$.
11 / 29
Avogadro’s number is approximately:
Reasoning: One mole contains $\mathbf{6.02 \times}\mathbf{10}^{\mathbf{23}}$ particles.
12 / 29
Boltzmann constant is numerically much smaller than $\mathbf{(R)}$ because:
Reasoning: Dividing $\mathbf{R}$ by the huge value of $\mathbf{N}_{\mathbf{A}}$ gives a very small constant.
13 / 29
Boltzmann constant relates temperature to:
Reasoning: It links temperature with the average energy of particles.
14 / 29
In $\mathbf{PV = NkT}$, increasing $\mathbf{(N)}$ while keeping $\mathbf{(V)}$ and $\mathbf{(T)}$ constant will:
Reasoning: Pressure is directly proportional to the number of molecules.
15 / 29
Which quantity has no unit?
Reasoning: Avogadro’s number is simply a count of particles.
16 / 29
If Avogadro’s number were larger, the value of $\mathbf{(k)}$ would be:
Reasoning: $\mathbf{k =}$ $\frac{\mathbf{R}}{\mathbf{N}_{\mathbf{A}}}$, so increasing $\mathbf{N}_{\mathbf{A}}$ decreases $\mathbf{(k)}$.
17 / 29
The quantity $\mathbf{kT}$ has units of:
Reasoning: $\mathbf{k}$ has units $\mathbf{JK}^{\mathbf{- 1}}$, so multiplying by $\mathbf{K}$ gives joules.
18 / 29
The Boltzmann constant is most useful when dealing with:
Reasoning: It is a microscopic constant used in molecular physics.
19 / 29
Which equation can be obtained by substituting $\mathbf{R =}\mathbf{kN}_{\mathbf{A}}$ into $\mathbf{PV = nRT\ }$?
Reasoning: Using $\mathbf{N =}\mathbf{nN}_{\mathbf{A}}$ leads directly to $\mathbf{PV = NkT}$.
20 / 29
The physical significance of $\mathbf{(k)}$ is that it:
Reasoning: Boltzmann constant bridges macroscopic and microscopic gas equations.
21 / 29
Which statement is correct?
Reasoning: $\mathbf{R = 8.31}$, while $\mathbf{k = 1.38 \times}\mathbf{10}^{\mathbf{- 23}}$.
22 / 29
The ratio $\frac{\mathbf{R}}{\mathbf{k}}$ equals:
Reasoning: Rearranging gives $\mathbf{N}_{\mathbf{A}}\mathbf{=}$ $\frac{\mathbf{R}}{\mathbf{k}}$.
23 / 29
Why is $\mathbf{PV = NkT}$ preferred in kinetic theory?
Reasoning: Kinetic theory deals with individual molecules.
24 / 29
The unit $\mathbf{JK}^{\mathbf{- 1}}$ suggests that Boltzmann constant relates:
Reasoning: The unit indicates energy per unit temperature.
25 / 29
One mole of gas contains $\mathbf{N}_{\mathbf{A}}$ molecules. Therefore:
Reasoning: Total molecules equal moles multiplied by Avogadro’s number.
26 / 29
If $\mathbf{N}_{\mathbf{A}}$ is substituted into $\mathbf{R =}\mathbf{kN}_{\mathbf{A\ }}$, then $\mathbf{R}$ represents:
Reasoning: The gas constant applies to one mole of gas.
27 / 29
The numerical value of $\mathbf{(k)}$ indicates that:
Reasoning: The small value reflects the tiny energy scale of individual molecules.
28 / 29
Which quantity is a fundamental constant of statistical physics?
Reasoning: Boltzmann constant is central to statistical mechanics.
29 / 29
A student forgets Avogadro’s number. Which expression can help recover it?
Reasoning: This follows directly from $\mathbf{k =}$ $\frac{\mathbf{R}}{\mathbf{N}_{\mathbf{A}}}$.
Your score is
Restart quiz