/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A \(0.35-\mathrm{m}^{3}\) vessel... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(0.35-\mathrm{m}^{3}\) vessel holds ethane vapor at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) and \(2200 \mathrm{kPa}\). If it is heated to \(493.15 \mathrm{K}\left(220^{\circ} \mathrm{C}\right),\) what pressure is developed?

Short Answer

Expert verified
The final pressure is 3641 kPa.

Step by step solution

01

Identify Given Variables

We have a vessel with volume \(V = 0.35 \text{ m}^{3}\), initial temperature \(T_1 = 298.15 \text{ K}\), initial pressure \(P_1 = 2200 \text{ kPa}\), and final temperature \(T_2 = 493.15 \text{ K}\). We need to find the final pressure \(P_2\).
02

Apply the Ideal Gas Law

Assume the gas behaves ideally. The Ideal Gas Law is \(PV = nRT\). Since \(n\) and \(V\) remain constant, we use the relation \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). This rearranges to \(P_2 = P_1 \times \frac{T_2}{T_1}\).
03

Substitute Known Values

Substitute the known values into the equation: \(P_2 = 2200 \text{ kPa} \times \frac{493.15 \text{ K}}{298.15 \text{ K}}\).
04

Calculate Final Pressure

Compute the result: \(P_2 = 2200 \times 1.655 = 3641 \text{ kPa}\).

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Key Concepts

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

Pressure Calculation
Calculating pressure in gases can be effortless once you grasp the fundamental concepts. One of the main equations involved is from the Ideal Gas Law: \( PV = nRT \). This relationship helps us find how changes in temperature and volume interact with pressure in a gas.

In scenarios where the number of moles \( n \) and volume \( V \) are constant, we utilize a simplified version of this equation: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). This equation tells us that the ratio of pressure to temperature remains the same. Since the variables \( n \) and \( R \) do not change, changes in pressure directly correlate with changes in temperature. In practice, this allows you to determine any of the unknowns as long as you have the others.
  • Start by identifying your known variables such as initial and final temperatures \( T_1 \) and \( T_2 \), initial pressure \( P_1 \), and ensure volume is constant.
  • Rearrange the equation to solve for the desired unknown, like \( P_2 \).
  • Substitute the known values into the equation and solve to find the final pressure. Calculations as demonstrated show that the final pressure can be swiftly computed given these relationships.
Temperature Conversion
Temperature conversion might seem straightforward, but it's critical in scientific contexts to ensure accuracy. The Ideal Gas Law uses Kelvin temperatures because Kelvin is the absolute temperature scale, meaning there are no negative values.

Many problems will initially provide temperature in Celsius, necessitating conversion to Kelvin for accurate gas law calculations. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
  • Start by determining the temperature in Celsius, which is often easier as an initial step.
  • Add 273.15 to the Celsius temperature to convert to Kelvin.
  • Use this Kelvin value in any Ideal Gas Law calculations to maintain consistency and accuracy.
For our exercise, the conversion ensures that the Ideal Gas Law applies correctly, allowing us to use initially 298.15 K \((25^{\circ} \text{C})\) and finally 493.15 K \((220^{\circ} \text{C})\).
Volume and Pressure Relationship
Understanding the relationship between volume and pressure is a pivotal aspect of gas behavior. Often, this is expressed through Boyle's Law, which describes how pressure varies inversely with volume at constant temperature in a closed system: \( P_1V_1 = P_2V_2 \).

However, when applying the Ideal Gas Law like in this exercise, both volume \( V \) and the number of gas moles \( n \) remain constant. Consequently, we explore how pressure responds solely to changes in temperature, as encapsulated by the equation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).
  • This type of problem, where volume is a fixed value, focuses on the temperature-pressure relationship.
  • Hence, there isn't a change in volume influencing the pressure development.
  • This allows learners to focus on how gas pressure increases with an increase in temperature, assuming volume is unchanged. This illustrates the direct relationship between pressure and temperature.
Knowing how these relationships behave makes it easier to predict changes in one variable when the others are altered.

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Most popular questions from this chapter

For a gas described by the Redlich/Kwong equation and for a temperature greater than \(T_{c},\) develop expressions for the two limiting slopes, $$\lim _{P \rightarrow 0}\left(\frac{\partial Z}{\partial P}\right)_{T} \quad \lim _{P \rightarrow \infty}\left(\frac{\partial Z}{\partial P}\right)_{T}$$ Note that in the limit as \(P \rightarrow 0, V \rightarrow \infty\), and that in the limit as \(P \rightarrow \infty, V \rightarrow \mathrm{b}\).

Calculate the reversible work done in compressing \(0.0283 \mathrm{m}^{3}\) of mercury at a constant temperature of \(273.15 \mathrm{K}\left(\mathrm{O}^{\circ} \mathrm{C}\right)\) from 1 atm to 3000 atm. The isothermal compressibility of mercury at \(273.15 \mathrm{K}\left(0^{\circ} \mathrm{C}\right)\) is $$\kappa=3.9 \times 10^{-6}-0.1 \times 10^{-9} P$$ where \(P\) is in atm and \(\mathrm{K}\) is in \(\mathrm{atm}^{-1}\).

The Tait equation for liquids is written for an isotherm as: $$V=V_{0}\left(1-\frac{A P}{B+P}\right)$$ where \(V\) is molar or specific volume, \(V_{0}\) is the hypothetical molar or specific volume at zero pressure, and \(\mathrm{A}\) and \(\mathrm{B}\) are positiveconstants. Find an expressionfor the isothermal compressibility consistent with this equation.

Natural gas (assume pure methane) is delivered to a city via pipeline at a volumetric rate of 4 normal \(\mathrm{Mm}^{3}\) per day. Average delivery conditions are \(283.15 \mathrm{K}\left(10^{\circ} \mathrm{C}\right)\) and 20.7 bar. Determine: (a) The volumetric delivery rate in actual \(m^{3}\) per day. (b) The molar delivery rate in kmol per hour. (c) The gas velocity at delivery conditions in \(\mathrm{m} \mathrm{s}^{-1}\) The pipe is \(600 \mathrm{mm}\) heavy duty steel with an inside diameter of \(575 \mathrm{mm}\). Normal conditions are \(273.15 \mathrm{K}\left(0^{\circ} \mathrm{C}\right)\) and \(1 \mathrm{atm}\).

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

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