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Question 1
A monatomic ideal gas contains $3.0 \times 1 0^{23}$ particles at a temperature of $400 K$ . What is the internal energy of the gas? (Use $k_{B} = 1.38 \times 1 0^{- 23} J K^{- 1}$ )
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $2.48 \times 1 0^{3} J$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the formula for internal energy
For a monatomic ideal gas, the internal energy is given by $U = \frac{3}{2} N k_{B} T$ where $N$ is the number of particles, $k_{B}$ is the Boltzmann constant, and $T$ is the absolute temperature.
Step 2: Substitute the given values
$U = \frac{3}{2} \times (3.0 \times 1 0^{23}) \times (1.38 \times 1 0^{- 23}) \times 400$
Step 3: Calculate step by step
First: $N k_{B} = 3.0 \times 1 0^{23} \times 1.38 \times 1 0^{- 23} = 4.14 J K^{- 1}$ . Then: $U = \frac{3}{2} \times 4.14 \times 400 = 1.5 \times 1656 = 2484 J \approx 2.48 \times 1 0^{3} J$ .
Step 4: Select the correct answer
The internal energy is approximately $2.48 \times 1 0^{3} J$ .
Method #2Approach 2
Step 1: Identify the calculation required
We need $U = \frac{3}{2} N k_{B} T$ . The expected order of magnitude: $N k_{B} \approx 3 \times 1 0^{23} \times 1.4 \times 1 0^{- 23} \approx 4.2$ , so $U \approx 1.5 \times 4.2 \times 400 \approx 2520 J$ .
Step 2: Eliminate $1.66 \times 10^3\,\text{J}$
This value corresponds to using $\frac{1}{2} N k_{B} T$ (using a factor of 1 instead of $\frac{3}{2}$ ) — this ignores that a monatomic gas has 3 translational degrees of freedom, not 1.
Step 3: Eliminate $3.31 \times 10^3\,\text{J}$
This is approximately double the expected value, perhaps from mistakenly using $\frac{3}{2} \times 2 \times N k_{B} T$ . It does not match the correct formula.
Step 4: Eliminate $4.97 \times 10^3\,\text{J}$
This is roughly $\frac{5}{2} N k_{B} T$ , which applies to diatomic (not monatomic) gases. The question specifies a monatomic gas.
Step 5: Select the correct answer
Only $2.48 \times 1 0^{3} J$ matches the correct application of $U = \frac{3}{2} N k_{B} T$ .
Question 2
A fixed amount of gas undergoes an isovolumetric (isochoric) process. Which of the following correctly describes this process?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $W = 0$ and $Δ U = Q$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Define an isochoric process
An isovolumetric (isochoric) process occurs at constant volume: $Δ V = 0$ .
Step 2: Determine work done
Since $W = P Δ V$ and $Δ V = 0$ , we have $W = 0$ . No work is done because the gas does not expand or compress.
Step 3: Apply the First Law
From $Δ U = Q - W$ with $W = 0$ : $Δ U = Q - 0 = Q$ All heat added to the gas goes directly into changing internal energy.
Step 4: Select the correct answer
The correct description is $W = 0$ and $ΔU = Q$ .
Method #2Approach 2
Step 1: Recall the four process types
We need to match the correct equations to an isochoric process. The First Law is $Δ U = Q - W$ .
Step 2: Eliminate '$Q = 0$ and $\Delta U = -W$'
This describes an adiabatic process, not isochoric. $Q = 0$ means no heat exchange — there is no requirement for this in an isochoric process.
Step 3: Eliminate '$\Delta U = 0$ and $Q = W$'
This describes an isothermal process where temperature (and hence internal energy) is constant. An isochoric process can have a changing temperature.
Step 4: Eliminate '$W = P\Delta V$ and $Q = \Delta U + W$'
This describes an isobaric process. While $Q = Δ U + W$ is just a rearrangement of the First Law, pairing it with $W = P Δ V$ implies $Δ V \neq = 0$ , which contradicts constant volume.
Step 5: Select the correct answer
Only ' $W = 0$ and $Δ U = Q$ ' correctly describes an isochoric process.

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14 more questions in this topic

Free preview 2/16

Question 1
A monatomic ideal gas contains $3.0 \times 1 0^{23}$ particles at a temperature of $400 K$ . What is the internal energy of the gas? (Use $k_{B} = 1.38 \times 1 0^{- 23} J K^{- 1}$ )
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $2.48 \times 1 0^{3} J$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the formula for internal energy
For a monatomic ideal gas, the internal energy is given by $U = \frac{3}{2} N k_{B} T$ where $N$ is the number of particles, $k_{B}$ is the Boltzmann constant, and $T$ is the absolute temperature.
Step 2: Substitute the given values
$U = \frac{3}{2} \times (3.0 \times 1 0^{23}) \times (1.38 \times 1 0^{- 23}) \times 400$
Step 3: Calculate step by step
First: $N k_{B} = 3.0 \times 1 0^{23} \times 1.38 \times 1 0^{- 23} = 4.14 J K^{- 1}$ . Then: $U = \frac{3}{2} \times 4.14 \times 400 = 1.5 \times 1656 = 2484 J \approx 2.48 \times 1 0^{3} J$ .
Step 4: Select the correct answer
The internal energy is approximately $2.48 \times 1 0^{3} J$ .
Method #2Approach 2
Step 1: Identify the calculation required
We need $U = \frac{3}{2} N k_{B} T$ . The expected order of magnitude: $N k_{B} \approx 3 \times 1 0^{23} \times 1.4 \times 1 0^{- 23} \approx 4.2$ , so $U \approx 1.5 \times 4.2 \times 400 \approx 2520 J$ .
Step 2: Eliminate $1.66 \times 10^3\,\text{J}$
This value corresponds to using $\frac{1}{2} N k_{B} T$ (using a factor of 1 instead of $\frac{3}{2}$ ) — this ignores that a monatomic gas has 3 translational degrees of freedom, not 1.
Step 3: Eliminate $3.31 \times 10^3\,\text{J}$
This is approximately double the expected value, perhaps from mistakenly using $\frac{3}{2} \times 2 \times N k_{B} T$ . It does not match the correct formula.
Step 4: Eliminate $4.97 \times 10^3\,\text{J}$
This is roughly $\frac{5}{2} N k_{B} T$ , which applies to diatomic (not monatomic) gases. The question specifies a monatomic gas.
Step 5: Select the correct answer
Only $2.48 \times 1 0^{3} J$ matches the correct application of $U = \frac{3}{2} N k_{B} T$ .
Question 2
A fixed amount of gas undergoes an isovolumetric (isochoric) process. Which of the following correctly describes this process?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $W = 0$ and $Δ U = Q$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Define an isochoric process
An isovolumetric (isochoric) process occurs at constant volume: $Δ V = 0$ .
Step 2: Determine work done
Since $W = P Δ V$ and $Δ V = 0$ , we have $W = 0$ . No work is done because the gas does not expand or compress.
Step 3: Apply the First Law
From $Δ U = Q - W$ with $W = 0$ : $Δ U = Q - 0 = Q$ All heat added to the gas goes directly into changing internal energy.
Step 4: Select the correct answer
The correct description is $W = 0$ and $ΔU = Q$ .
Method #2Approach 2
Step 1: Recall the four process types
We need to match the correct equations to an isochoric process. The First Law is $Δ U = Q - W$ .
Step 2: Eliminate '$Q = 0$ and $\Delta U = -W$'
This describes an adiabatic process, not isochoric. $Q = 0$ means no heat exchange — there is no requirement for this in an isochoric process.
Step 3: Eliminate '$\Delta U = 0$ and $Q = W$'
This describes an isothermal process where temperature (and hence internal energy) is constant. An isochoric process can have a changing temperature.
Step 4: Eliminate '$W = P\Delta V$ and $Q = \Delta U + W$'
This describes an isobaric process. While $Q = Δ U + W$ is just a rearrangement of the First Law, pairing it with $W = P Δ V$ implies $Δ V \neq = 0$ , which contradicts constant volume.
Step 5: Select the correct answer
Only ' $W = 0$ and $Δ U = Q$ ' correctly describes an isochoric process.

Free preview

Step-by-step walkthrough

Step 1: Identify the formula for internal energy

Step 2: Substitute the given values

Step 3: Calculate step by step

Step 4: Select the correct answer

Step 1: Identify the calculation required

Step 2: Eliminate $1.66 \times 10^3\,\text{J}$

Step 3: Eliminate $3.31 \times 10^3\,\text{J}$

Step 4: Eliminate $4.97 \times 10^3\,\text{J}$

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Define an isochoric process

Step 2: Determine work done

Step 3: Apply the First Law

Step 4: Select the correct answer

Step 1: Recall the four process types

Step 2: Eliminate '$Q = 0$ and $\Delta U = -W$'

Step 3: Eliminate '$\Delta U = 0$ and $Q = W$'

Step 4: Eliminate '$W = P\Delta V$ and $Q = \Delta U + W$'

Step 5: Select the correct answer

14 more questions in this topic

Step-by-step walkthrough

Step 1: Identify the formula for internal energy

Step 2: Substitute the given values

Step 3: Calculate step by step

Step 4: Select the correct answer

Step 1: Identify the calculation required

Step 2: Eliminate $1.66 \times 10^3\,\text{J}$

Step 3: Eliminate $3.31 \times 10^3\,\text{J}$

Step 4: Eliminate $4.97 \times 10^3\,\text{J}$

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Define an isochoric process

Step 2: Determine work done

Step 3: Apply the First Law

Step 4: Select the correct answer

Step 1: Recall the four process types

Step 2: Eliminate '$Q = 0$ and $\Delta U = -W$'

Step 3: Eliminate '$\Delta U = 0$ and $Q = W$'

Step 4: Eliminate '$W = P\Delta V$ and $Q = \Delta U + W$'

Step 5: Select the correct answer

14 more questions in this topic