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Chapter 3: Problem 23

One kmol of an ideal gas, initially at \(303.15 \mathrm{K} / 30^{\circ} \mathrm{C}\) )and \(/\) bar, undergoes the following mechanically reversible changes. It is compressed isothermally to a point such that when it is heated at constant volume to \(393.15 \mathrm{K}\left(120^{\circ} \mathrm{C}\right)\) its final pressure is 12 bar. Calculate \(Q, W, \Delta U,\) and \(A H\) for the process. Take \(C_{p}=(7 / 2) R\) and \(C_{V}=(5 / 2) R\).

Short Answer

Expert verified
Calculate 螖U using \( \Delta U = nC_V\Delta T \), find Q using first law, W from isothermal work, and 螖H using \( \Delta H = nC_p\Delta T \).

Step by step solution

01

Determine Initial Conditions

We start by identifying the initial conditions: The ideal gas is at 303.15 K and 1 bar. We know that the number of moles, \( n \), is 1 kmol.
02

Calculate Isothermal Compression Work (W)

The work done during an isothermal process is calculated using the equation \( W = nRT \ln\left(\frac{P_2}{P_1}\right) \). Given \( T = 303.15 \) K and \( P_1 = 1 \) bar, we need to find \( P_2 \) for the final temperature of 393.15 K and final pressure of 12 bar to solve for \( P_{2,1} \) in the isothermal process.
03

Calculate Final Volume Change to Obtain P2

Using \( P_2 = \frac{nRT_2}{V_2} \) and the fact that \( P_2 = 12 \text{ bar} \) at the end of the constant volume process, solve \( V_2 \) using initial condition P-V-T information after substituting \( V_1 \) from the first process.
04

Find Work Done During Heating at Constant Volume

Since the volume is constant, the work done, \( W \), is zero during this part of the process.
05

Calculate Changes in Internal Energy (螖U)

For any process, the change in internal energy \( \Delta U \) can be calculated using \( \Delta U = nC_V\Delta T \). Substituting \( n = 1 \), \( C_V = \frac{5}{2}R \), and \( \Delta T = T_2 - T_1 = 393.15 \text{ K} - 303.15 \text{ K} \), solve for \( \Delta U \).
06

Determine Heat Transfer (Q)

Use the first law of thermodynamics, \( \Delta U = Q - W \). Since \( \Delta U \) and \( W \) have been calculated, solve for \( Q \).
07

Compute Change in Enthalpy (螖H)

The change in enthalpy is calculated using \( \Delta H = nC_p\Delta T \), where \( n = 1 \), \( C_p = \frac{7}{2}R \), and \( \Delta T = 393.15 - 303.15 \) K. Substitute these values to find \( \Delta H \).

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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 linking pressure, volume, and temperature of an ideal gas. The equation is given by: \[ PV = nRT \] Here, \( P \) is the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvins.
  • Helps relate and predict the state of an ideal gas.
  • Assumes ideal conditions, where gas particles have no volume and no interactions.
In real-world scenarios, it offers a good approximation, particularly at high temperatures and low pressures. During isothermal compression, we use this law to find initial and final pressures and volumes.
Isothermal Process
An isothermal process maintains constant temperature throughout the thermodynamic process. This is achieved by making sure heat exchange occurs between the system and surroundings, keeping temperature stable.
  • The equation used is \( W = nRT \ln\left(\frac{P_2}{P_1}\right) \) for work done.
  • Involves energy transfer but no internal energy change since temperature remains the same.
During such processes, computations are simplified due to the constant temperature. This process aids in understanding how work can be extracted or required during isothermal expansions or compressions.
Internal Energy
The Internal Energy of a system refers to the total energy contained within a system due to molecular motion and interaction. It's exclusively a function of temperature for ideal gases.
  • For an ideal gas, it depends only on temperature, not on pressure or volume.
  • The change, \( \Delta U \), is given by \( nC_V\Delta T \), reflecting changes due to heating or cooling.
In scenarios like constant volume heating, there's no work involved in volume change. Hence, any energy change translates to a direct rise in internal energy. Understanding this helps clarify energy flow within gaseous systems when subject to processes like expansion or compression.
Enthalpy
Enthalpy is a measurement of total heat content in a thermodynamic system. For processes conducted under constant pressure, it serves as an insightful parameter.
  • The change in enthalpy is mainly used to understand heat exchange in processes.
  • The formula \( \Delta H = nC_p\Delta T \) is used for such calculations, especially relevant during heating or cooling.
In the discussed exercise, enthalpy provides insights into how much heat is transferred even when pressure conditions remain unchanged. It's critical to correlate such changes with both temperature variations and heat capacity of the gas to grasp overall thermodynamic behavior.

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Most popular questions from this chapter

Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which the temperature changes from \(273.15 \mathrm{K}\) \(\left(0^{\circ} \mathrm{C}\right)\) to \(293.15 \mathrm{K}\left(20^{\circ} \mathrm{C}\right) .\) Determine \(\Delta V^{t}, \mathrm{W}, \mathrm{Q}, A H^{\prime},\) and \(\Delta U^{\prime} .\) The properties for liquid carbon tetrachloride at 1 bar and \(273.15 \mathrm{K}\left(0^{\circ} \mathrm{C}\right)\) may be assumed independent of temperature: \(\beta=1.2 \times 10^{-3} \mathrm{K}^{-1}, C_{P}=0.84 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{K}^{-1},\) and \(\rho=1590 \mathrm{kg} \mathrm{m}^{-3}\).

A process consists of two steps: (1) One \(\mathrm{kmol}\) of air at \(\mathrm{T}=800 \mathrm{K}\) and \(P=4\) bar is cooled at constant volume to \(\mathrm{T}=350 \mathrm{K}\). ( 2 ) The air is then heated at constant pressure until its temperature reaches \(800 \mathrm{K}\). If this two-step process is replaced by a single isothermal expansion of the air from \(800 \mathrm{K}\) and 4 bar to some final pressure \(\mathrm{P}\), what is the value of \(P\) that makes the work of the two processes the same? Assume mechanical reversibility and treat air as an ideal gas with \(C_{P}=(7 / 2) R\) and \(C_{V}=(5 / 2) R\).

Express the volume expansivity and the isothermal compressibility as functions of density \(\rho\) and its partialderivatives.For water at \(323.15 \mathrm{K}\left(50^{\circ} \mathrm{C}\right)\) and 1 bar, \(\mathrm{K}=44.18 \times 10^{-6}\) bar \(^{-1}\). To what pressure must water be compressed at \(323.15 \mathrm{K}\left(50^{\circ} \mathrm{C}\right)\) to change its density by \(1 \%\) ? Assume that \(\kappa\) is independent of \(\mathbf{P}\).

Generally, volume expansivity \(\beta\) and isothermal compressibility \(\kappa\) depend on \(\mathrm{T}\) and \(\mathrm{P}\). Prove that: $$\left(\frac{\partial \beta}{\partial P}\right)_{T}=-\left(\frac{\partial \kappa}{\partial T}\right)_{P}$$

An ideal gas flows through a horizontal tube at steady state. No heat is added and no shaft work is done. The cross-sectional area of the tube changes with length, and this causes the velocity to change. Derive an equation relating the temperatureto the velocity of the gas. If nitrogen at \(423.15 \mathrm{K}\left(150^{\circ} \mathrm{C}\right)\) flows past one section of the tube at a velocity of \(2.5 \mathrm{m} \mathrm{s}^{-1},\) what is its temperature at another section where its velocity is \(50 \mathrm{m} \mathrm{s}^{-1} ?\) Let \(C_{P}=(7 / 2) R\).
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