Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 23

Calculate the amount of heat needed to raise the temperature of \(0.325 \mathrm{~mol}\) of a monatomic idcal gas from a temperature of \(-25^{\circ} \mathrm{C}\) to \(+50^{\circ} \mathrm{C}\) at (a) constant volume and (b) constant pressure. (c) Why is more heat needed to raise the temperature at constant pressure than at constant volume?

Short Answer

Expert verified
The heat needed at constant volume is approximately 3033 J, and at constant pressure, it's approximately 5056 J. More heat is needed at constant pressure because some of the heat is used to do work on the surroundings in addition to raising the temperature.

Step by step solution

01

Convert temperatures to Kelvin

To calculate the heat needed to change the temperature of a gas, we must express temperatures in Kelvin. Convert -25°C to Kelvin by adding 273.15, resulting in 248.15 K. Convert 50°C to Kelvin by adding 273.15, resulting in 323.15 K.
02

Calculate the change in temperature

Subtract the initial temperature in Kelvin from the final temperature in Kelvin to find the change in temperature: \(\text{Temperature change} = 323.15 K - 248.15 K = 75 K.\)
03

Calculate heat at constant volume (a)

To calculate the heat (q) at constant volume, use the formula \(q = nC_V\text{Temperature change}\), where \(n\) is the number of moles, and \(C_V\) is the molar heat capacity at constant volume for a monatomic ideal gas. We know that \(C_V = \frac{3}{2}R\), where \(R=8.3145 J/(mol \times K)\). Using \(0.325 mol\) of the gas, calculate \(q\) at constant volume: \(q = 0.325 \times \frac{3}{2} \times 8.3145 \times 75\).
04

Calculate heat at constant pressure (b)

To calculate the heat (q) at constant pressure, use the formula \(q = nC_P\text{Temperature change}\), where \(C_P\) is the molar heat capacity at constant pressure for a monatomic ideal gas. We know that \(C_P = \frac{5}{2}R\). Using \(0.325 mol\) of the gas, calculate \(q\) at constant pressure: \(q = 0.325 \times \frac{5}{2} \times 8.3145 \times 75\).
05

Explain the difference in heat needed (c)

More heat is required at constant pressure because the gas does work on its surroundings as it expands, in addition to the energy required to increase its internal energy. At constant volume, there is no work done since the volume is not allowed to change, so all the heat goes into increasing internal energy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Molar Heat Capacity
Understanding molar heat capacity is fundamental when dealing with thermodynamics heat calculations. This property tells us how much heat energy is required to raise the temperature of one mole of a substance by one degree Celsius (or one Kelvin). It's denoted as C and varies depending on whether the process occurs at constant volume (CV) or constant pressure (CP).

In our example, we use specific values for a monatomic ideal gas, where CV is \(\frac{3}{2}R\), and CP is \(\frac{5}{2}R\). The gas constant R is 8.3145 J/(mol·K). Thus, knowing the molar heat capacity is pivotal to calculate the heat needed for a temperature change.
Ideal Gas Law
The ideal gas law is a cornerstone of thermodynamics, represented by the equation PV = nRT. It relates the pressure (P), volume (V), the number of moles (n), temperature (T) in Kelvin, and the ideal gas constant (R).

While the ideal gas law is not directly used for heat calculations in this particular problem, it forms the background understanding for why different forms of heat capacity (CV versus CP) exist and how they are derived.
Temperature Conversion
Temperature plays a vital role in thermodynamics, and its measurement needs consistency. Most thermodynamics formulas require temperature in Kelvin. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. In this exercise, temperatures were converted from Celsius to Kelvin as the first step because all subsequent calculations for the heat required rely on the Kelvin scale.
Heat at Constant Volume
Calculating heat at constant volume is a key concept when a substance undergoes a temperature change without any variation in volume. The formula used is q = nCVΔT, with q representing the heat energy, n the number of moles, CV the molar heat capacity at constant volume, and ΔT the temperature change in Kelvin. No work is done on the surroundings in this case; all the heat goes into raising the temperature.
Heat at Constant Pressure
On the other hand, heat at constant pressure involves conditions where the volume can change while the system is heated. The pertinent formula is q = nCPΔT. The molar heat capacity at constant pressure CP is used because work can be done by the system as it expands against the external pressure. In our example, it's why more heat is required at constant pressure compared to constant volume—it accounts for both the increase in internal energy and the work done by the system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the enthalpy of the reaction \(\mathrm{P}_{4}(\mathrm{~s})+10 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{~s})\) from the reactions $$ \begin{gathered} \mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{3}(\mathrm{I}) \\ \Delta H^{\circ}=-1278.8 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{~s}) \\ \Delta H^{m}=-124 \mathrm{~kJ} \end{gathered} $$

A piece of metal of mass \(20.0 \mathrm{~g}\) at \(100.0^{\circ} \mathrm{C}\) is placed in a calorimeter containing \(50.7 \mathrm{~g}\) of watcr at \(22.0^{\circ} \mathrm{C}\). The final temperature of the mixture is \(25.7^{\circ} \mathrm{C}\) What is the specific heat capacity of the mctal? Assume that all the energy lost by the metal is gained by the water.

(a) Near room temperature the specific heat capacity of ethanol is \(2.42 \mathrm{~J} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1} \cdot \mathrm{g}^{-1}\). Calculate the heat that must be removed to reduce the temperature of \(150.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\) from \(50.0^{\circ} \mathrm{C}\) to \(16.6^{\circ} \mathrm{C}\). (b) What mass of copper can be heated from \(15^{\circ} \mathrm{C}\) to \(205^{\circ} \mathrm{C}\) when \(425 \mathrm{~kJ}\) of energy is available?

When \(25.0 \mathrm{~g}\) of a metal at a temperature of \(90.0^{\circ} \mathrm{C}\) is added to \(50.0 \mathrm{~g}\) of water at \(25.0^{\circ} \mathrm{C}\), the water temperature rises to \(29.8^{\circ} \mathrm{C}\). The specific heat capacity of water is \(4.184 \mathrm{~J}-\left({ }^{\circ} \mathrm{C}\right)^{-1} \mathrm{~g}^{-1}\). What is the specific heat capacity of the metal?

Near room temperature, the specific heat capacity of benzene is \(1.05 \mathrm{~J} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1} \mathrm{~g}^{-1}\). Calculate the heat nceded to raise the temperature of \(50.0 \mathrm{~g}\) of benzene from \(25.3^{\circ} \mathrm{C}\) to \(37.2^{\circ} \mathrm{C}\). (b) A \(1.0-\mathrm{kg}\) block of aluminum is supplied with \(490 \mathrm{~kJ}\) of heat. What is the temperature change of the aluminum? The specific heat capacity of aluminum is \(0.90 \mathrm{~J} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1} \cdot \mathrm{g}^{-1}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free