Under constant pressure P = 18 bar an ideal gas of 1.5 mol undergoes expansion from 1.2L to 23.5L, and Cv,m 20.79J/mol/K. Please calculate q, w, AU, and AH Instruction: = (1) all units are KJ (not J); (2) Please round to 1 decimal place (not E notation). For example: 23.010 should be entered as 23.0; If the answer is negative, please add the negative sign. If answer is zero, please just enter 0 without decimal. (1) w= (2) q= (3) AU= (4) ΔΗ=
Under constant pressure P = 18 bar an ideal gas of 1.5 mol undergoes expansion from 1.2L to 23.5L, and Cv,m 20.79J/mol/K. Please calculate q, w, AU, and AH Instruction: = (1) all units are KJ (not J); (2) Please round to 1 decimal place (not E notation). For example: 23.010 should be entered as 23.0; If the answer is negative, please add the negative sign. If answer is zero, please just enter 0 without decimal. (1) w= (2) q= (3) AU= (4) ΔΗ=
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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can u help, the 2nd pic is an example of how to solve
![**QUESTION 1**
Under constant pressure \( P = 18 \) bar, an ideal gas of 1.5 mol undergoes expansion from 1.2 L to 23.5 L, and \( C_{V,m} = 20.79 \) J/mol/K. Please calculate \( q, w, \Delta U, \) and \( \Delta H \).
**Instruction:**
1. All units are kJ (not J).
2. Please round to 1 decimal place (not E notation). For example: 23.010 should be entered as 23.0; If the answer is negative, please add the negative sign. If the answer is zero, please just enter 0 without a decimal.
(1) \( w = \) [ ]
(2) \( q = \) [ ]
(3) \( \Delta U = \) [ ]
(4) \( \Delta H = \) [ ]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4dfc165-1571-419f-99fc-a4587d76c340%2F12bc028c-74b7-44b0-8772-f3c499cd383e%2Fk5oi10q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**QUESTION 1**
Under constant pressure \( P = 18 \) bar, an ideal gas of 1.5 mol undergoes expansion from 1.2 L to 23.5 L, and \( C_{V,m} = 20.79 \) J/mol/K. Please calculate \( q, w, \Delta U, \) and \( \Delta H \).
**Instruction:**
1. All units are kJ (not J).
2. Please round to 1 decimal place (not E notation). For example: 23.010 should be entered as 23.0; If the answer is negative, please add the negative sign. If the answer is zero, please just enter 0 without a decimal.
(1) \( w = \) [ ]
(2) \( q = \) [ ]
(3) \( \Delta U = \) [ ]
(4) \( \Delta H = \) [ ]
![**Transcription of Thermodynamic Calculations**
### Step 2
The change in internal energy during the process is calculated as follows:
\[
\Delta U = n C_v (T_2 - T_1) = 1.5 \, \text{mol} \times \frac{3}{2} \times 8.314 \, \frac{J}{\text{mol} \cdot K} \times (303 - 393) \, K
\]
\[
= -1683.585 \, J
\]
\[
\Delta U \, (J) = -1.68 \times 10^3
\]
### Step 3
For an ideal gas, we have the relationships:
- \(C_p - C_v = R\)
- \(C_p = 5R/2 \, (R + 3R/2)\)
The change in enthalpy is calculated as follows:
\[
\Delta H = n C_p (T_2 - T_1) = 1.5 \, \text{mol} \times \frac{5}{2} \times 8.314 \, \frac{J}{\text{mol} \cdot K} \times (303 - 393) \, K
\]
\[
= -2805.975 \, J
\]
\[
\Delta H \, (J) = -2.81 \times 10^3
\]
For a constant pressure process, the heat transfer is equal to the change in internal energy. Thus:
\[
q \, (J) = -2.81 \times 10^3
\]
### Step 4
The work done during the process is calculated by applying the first law of thermodynamics as follows:
\[
q - w = \Delta U
\]
\[
-2.81 \times 10^3 - w = -1.68 \times 10^3
\]
\[
w = 1.13 \times 10^3
\]
\[
w \, (J) = -1.13 \times 10^3
\]
This sequence of calculations demonstrates how changes in internal energy and enthalpy are derived for a thermodynamic process involving an ideal gas.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4dfc165-1571-419f-99fc-a4587d76c340%2F12bc028c-74b7-44b0-8772-f3c499cd383e%2Fzlxc0y_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription of Thermodynamic Calculations**
### Step 2
The change in internal energy during the process is calculated as follows:
\[
\Delta U = n C_v (T_2 - T_1) = 1.5 \, \text{mol} \times \frac{3}{2} \times 8.314 \, \frac{J}{\text{mol} \cdot K} \times (303 - 393) \, K
\]
\[
= -1683.585 \, J
\]
\[
\Delta U \, (J) = -1.68 \times 10^3
\]
### Step 3
For an ideal gas, we have the relationships:
- \(C_p - C_v = R\)
- \(C_p = 5R/2 \, (R + 3R/2)\)
The change in enthalpy is calculated as follows:
\[
\Delta H = n C_p (T_2 - T_1) = 1.5 \, \text{mol} \times \frac{5}{2} \times 8.314 \, \frac{J}{\text{mol} \cdot K} \times (303 - 393) \, K
\]
\[
= -2805.975 \, J
\]
\[
\Delta H \, (J) = -2.81 \times 10^3
\]
For a constant pressure process, the heat transfer is equal to the change in internal energy. Thus:
\[
q \, (J) = -2.81 \times 10^3
\]
### Step 4
The work done during the process is calculated by applying the first law of thermodynamics as follows:
\[
q - w = \Delta U
\]
\[
-2.81 \times 10^3 - w = -1.68 \times 10^3
\]
\[
w = 1.13 \times 10^3
\]
\[
w \, (J) = -1.13 \times 10^3
\]
This sequence of calculations demonstrates how changes in internal energy and enthalpy are derived for a thermodynamic process involving an ideal gas.
Expert Solution
Step 1
Given data:
P = 18 bar
n = 1.5 mol
V1 = 1.2 L
V2 = 23.5 L
Step by step
Solved in 4 steps
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