The initial state of \(0.1\) mol of an ideal monatomic gas is \(P_{0}=32 \mathrm{~Pa}\) and \(V_{0}=8 \mathrm{~m}^{3}\). The final state is \(P_{1}=1 \mathrm{~Pa}\) and \(V_{1}=64 \mathrm{~m}^{3}\). Suppose that the gas undergoes a process along a straight line joining these two states with an equation \(P=a V+b\), where \(a=-31 / 56\) and \(b=255 / 7\). Plot this straight line to scale on a \(P V\) diagram. Calculate: (a) Temperature \(T\) as a function of \(V\) along the straight line. (b) The value of \(V\) at which \(T\) is a maximum. (c) The values of \(T_{0}, T_{\max }\), and \(T_{1}\). (d) The heat \(Q\) transferred from the volume \(V_{0}\) to any other volume \(V\) along the straight line. (e) The values of \(P\) and \(V\) at which \(Q\) is a maximum. (f) The heat transferred along the line from \(V_{0}\) to \(V\) when \(Q\) is a maximum. (g) The heat transferred from \(V\) at maximum \(Q\) to \(V_{1}\).

Step by step solution

01

- Understand the equation of the process

The gas undergoes a process represented by the equation: \[ P = aV + b \] with given constants \( a = -\frac{31}{56} \) and \( b = \frac{255}{7} \).

02

- Express Temperature T as a Function of V

Use the ideal gas law \( PV = nRT \) to write temperature as: \[ T = \frac{PV}{nR} \] Substitute \( P \) from the line equation: \[ T = \frac{(aV + b)V}{nR} \]

03

- Find the Volume V at Maximum Temperature

To find \( V \) for maximum \( T \), take the derivative of \( T \) with respect to \( V \) and set to zero: \[ \frac{dT}{dV} = 0 \] Given \[ T(V) = \frac{(aV^2 + bV)}{nR} \] Differentiate: \[ \frac{d}{dV} \left( \frac{aV^2 + bV}{nR} \right) = 0 \] Simplify to find \( V \): \[ \frac{2aV + b}{nR} = 0 \] \[ 2aV + b = 0 \] Solve for \( V \): \[ V = -\frac{b}{2a} \]

04

- Calculate the values of T₀, Tₘₐₓ, and T₁

Calculate initial \((T_0)\), maximum \((T_{max})\), and final \((T_1)\) temperatures using the equation for temperature: \[ T_0 = \frac{P_0 V_0}{nR} \] \[ T_{max} = \frac{P(V) V}{nR} \] at maximum volume \( V \) calculated in step 3 \[ T_1 = \frac{P_1 V_1}{nR} \]

05

- Determine the Heat Q transferred

To find the heat \( Q \) transferred, use the first law of thermodynamics: \[ \Delta U = Q - W \] For monatomic gas: \( \Delta U = \frac{3}{2}nR\Delta T \) Work done \( W \): \[ W = \int_{V_0}^{V} P \, dV = \int_{V_0}^{V} (aV+b) \, dV \] Solve for \( Q \).

06

- Values of P and V at maximum Q

Maximum \( Q \) is where the work done is maximum, which happens at the transition points. Solve the derivative \( \frac{dQ}{dV} = 0 \).

07

- Heat transferred at maximum Q

Evaluate \( Q \) using the calculated values of \( P \) and \( V \) from the previous step.

08

- Heat transferred from V at maximum Q to V₁

Integrate \( PV \) equation again from maximum \( V \) to final \( V \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics and is given by: \( PV = nRT \). It relates the pressure \( P \), volume \( V \), and temperature \( T \) of an ideal gas with the amount of the gas in moles \( n \) and the gas constant \( R \). This law is crucial as it helps us understand the behavior of gases under different conditions.
In our exercise, we use the ideal gas law to express temperature \( T \) as a function of volume \( V \). By substituting the process equation \( P = aV + b \) into the ideal gas law, we get:

• \( T = \frac{(aV + b)V}{nR} \)

This equation shows how temperature changes with volume during the process. It is essential in calculations for finding maximum temperatures and other properties.

Thermodynamic Processes

In thermodynamics, different processes describe how a system changes from one state to another. Our exercise involves a process along a straight line joining two states of an ideal monatomic gas:

• Initial state: \( P_0 = 32 \, \text{Pa} \) and \( V_0 = 8 \, \text{m}^3 \)
• Final state: \( P_1 = 1 \, \text{Pa} \) and \( V_1 = 64 \, \text{m}^3 \)

The process can be represented as a linear equation \( P = aV + b \) with given constants \( a \) and \( b \). Understanding this process helps us calculate various properties like temperature maxima and heat transfer.
Thermodynamic processes can be:

• Isothermal (constant temperature)
• Adiabatic (no heat exchange)
• Isobaric (constant pressure)
• Isochoric (constant volume)

The process in our exercise is not one of these standard types but follows a unique path defined by the equation.

Heat Transfer Calculation

Heat transfer in thermodynamics involves the transfer of energy from one place or system to another. It's governed by the first law of thermodynamics: \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.
For our monatomic ideal gas:

• Internal energy change, \( \Delta U = \frac{3}{2}nR \Delta T \)

To calculate \( Q \), we need the work done \( W \):
\[ W = \int_{V_0}^{V} P \, dV = \int_{V_0}^{V} (aV+b) \, dV \]
By solving this, we can find the heat transfer as:

• \( Q = \Delta U + W \)

This is how we quantify the heat transferred during the process.

Temperature as Function of Volume

In the given exercise, temperature \( T \) varies as a function of volume \( V \) along the straight line process:
\[ T = \frac{(aV + b)V}{nR} \]
This equation derives from combining the ideal gas law and the process equation. To find the volume at which temperature is maximum, we take the derivative of \( T \) with respect to \( V \) and set it to zero:
\[ \frac{dT}{dV} = 0 \]
Simplifying gives:
\[ 2aV + b = 0 \]

• The volume at maximum temperature, \( V = -\frac{b}{2a} \)

By calculating the maximum temperature and verifying with initial and final states, we obtain the complete temperature profile through the process.