### Understanding the Thermodynamic Process of an Ideal Gas#### Problem StatementA sample of an ideal gas goes through the process shown in the figure below. 1. From A to B, the process is adiabatic.2. From B to C, it is isobaric with 345 kJ of energy entering the system by heat.3. From C to D, the process is isothermal.4. From D to A, it is isobaric with 371 kJ of energy leaving the system by heat.Determine the difference in internal energy \( E_{\text{Int,B}} - E_{\text{Int,A}} \).#### Given Data- Energy entering from B to C: \(345\ \text{kJ}\)- Energy leaving from D to A: \(371\ \text{kJ}\)#### Diagram DescriptionThe diagram displays a Pressure (P) vs. Volume (V) graph with the following key points:- **Point A:** \( P = 1\ \text{atm}, V = 0.09\ \text{m}^3 \)- **Point B:** \( P = 3\ \text{atm}, V = 0.2\ \text{m}^3 \)- **Point C:** \( P = 3\ \text{atm}, V = 0.4\ \text{m}^3 \)- **Point D:** \( P = 1\ \text{atm}, V = 1.2\ \text{m}^3 \)The graph shows the transitions among these points:- **A to B:** An upward curved line indicating an adiabatic process.- **B to C:** A horizontal line indicating an isobaric process.- **C to D:** A downward curved line indicating an isothermal process.- **D to A:** A horizontal line indicating an isobaric process.\( \Delta E_{\text{Int}} = E_{\text{Int,B}} - E_{\text{Int,A}} \)By examining these processes and the energy exchanges, we aim to calculate the change in internal energy \(\Delta E_{\text{Int}} \).#### CalculationWe apply the first law of thermodynamics to each process segment and then integrate the results to find the change in internal energy. ``` First law of thermodynamics:ΔU = Q - WWhere

Introduction: The gases which follows the Boyle's law, Charles's law, Avogadro's law and…

A sample of an ideal gas goes through the process shown in the figure below. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 k of energy entering the system by heat; from C to D, the process is isothermal; and from D to A, it is isobaric with 371 k of energy leaving the system by heat. Determine the difference in internal energy Ent, - Ent. A P (atm)

A sample of an ideal gas goes through the process shown in the figure below. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 k of energy entering the system by heat; from C to D, the process is isothermal; and from D to A, it is isobaric with 371 k of energy leaving the system by heat. Determine the difference in internal energy Ent, - Ent. A P (atm)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

A closed cylinder contains 1.0 mole of ideal monatomic gas at 300 oC and 4.0 bar.Insulating the cylinder so that there is no heat exchange with the surroundings, the gas isexpanded to 200 oC and 1.0 bar producing 1247 J of work.a) Is this process reversible or irreversible?b) What is the change in enthalpy (∆H) of the gas?c) What is the change in internal energy (∆U) of thegas?
For both circumstances (reversible and irreversible) what is the change in enthalpy dS for the gas, the surroundings (assuming ideal isothermal bath), and the universe?
For the reaction P4010(s) + 6 H20(1) 4 H3PO4(aq) AG° = -448.9 kJ and AS° = -16.3 J/K at 261 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 261 K. The standard enthalpy change for the reaction of 2.15 moles of P4010(s) at this temperature would be kJ. Submit Answer Try Another Version 1 item attempt remaining rch a
PART 4 PLEASE
Q1/ Calculate the final pressure of a sample of water vapor that expands reversibly and adiabatically from 24.5 Torr and 0.5 dm' to 3 dm"? Take Cp-7.3 J.K', Cv-3 J.K'
3. Derive the thermodynamic equation of state for an ideal gas starting from internal energy then express this equation in a form without differentials. = -P +T| T LƏT.
A monatomic ideal gas expands in a reversible adiabatic process. The initial temperature of the gas amounts to T; and the mole number is equal to n. The final temperature T; = T/4. Calculate the work done by the gas? a. -5/8 nRT; O b. -9/8 nRT; O c. 3/2 nRT; O d. 5/2 nRT O e. 9/8 nRT;
Example (2). 1 g of water at 273°K is converted into steam at the same temperature. The volume of water becomes 1671 ml on boiling. Calculate the change in the internal energy of the system if the heat of vaporization is 540 cal/g.
4. Calculate the work done under isothermal conditions for 5.5 mols of an ideal gas expanding from 10.0 bar to 2.00 bar at –20.0°C both for a reversible (a) and irreversible (b) process. For the latter, Pext = 2.00 bar. The general expression for work is, w=-[P_dV
Consider the expression: dS=dT - VadP. Suppose that water has ß = 4.53x10-5 atm¯¹, V = 18 cm³/mol, C = 18 cal/deg-mol, and a = 2.0 x10 deg¹. Compute the decrease in temperature which occurs if water at 25 °C and 1000 atm pressure is brought reversibly and adiabatically to 1 atm pressure.
Knowing that the molar heat capacity at constant pressure of CO₂ has the form Cp,m = (a + b + c/T²) J mol-¹K-¹, where a= 44.22, b= 8.79 x 10-³ K-¹, and c= -8.62 x 105 K 2,calculate how much heat can absorb 1.5 moles of this gas when the temperature is increased from 25°C to 150°C. NOTE: Give your answer in kJ mol-1
81.6 g of a metal alloy was heated to 93 o C, then plunged into 105 mL of H 2O that was at 19.5 o C. The system equilibrated at 22.3 oC. What is the specific heat of the metal?