A closed, non-conducting, horizontal cylinder is fitted with a non-conducting frictionless,floating piston which divides the cylinder into sections A and B. The two sections containequal masses of air, initially at the same conditions, T₁ = 300K, and P₁=1 atm. Anelectrical heating element in section A is activated, and the air temperatures slowlyincreases: TA in section A because of heat transfer, and TB in section B because ofadiabatic compression by the slowly moving piston. Treat air as an ideal gas with Cp =7/2R, and let na be the number of moles of air in section A. If the final temperature insection B (TB) is 325K, evaluate:a. TA, the temperature in section Ab. Q/nA, the heat absorbed per mole of Ac. P(final), the final pressured. The change in entropy (ASsys) of each section.

Answered: A closed, non-conducting, horizontal…

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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Question

A closed, non-conducting, horizontal cylinder is fitted with a non-conducting frictionless,
floating piston which divides the cylinder into sections A and B. The two sections contain
equal masses of air, initially at the same conditions, T₁ = 300K, and P₁=1 atm. An
electrical heating element in section A is activated, and the air temperatures slowly
increases: TA in section A because of heat transfer, and TB in section B because of
adiabatic compression by the slowly moving piston. Treat air as an ideal gas with Cp =
7/2R, and let na be the number of moles of air in section A. If the final temperature in
section B (TB) is 325K, evaluate:
a. TA, the temperature in section A
b. Q/nA, the heat absorbed per mole of A
c. P(final), the final pressure
d. The change in entropy (ASsys) of each section.

Expert Solution

Step 1

Given that-

For a closed system divided into sections A and B.

Mass of air in sections A and B are equal so as moles are also equal.

Hence, n_A=n_B

Initial tempearture and pressure for both the systems are similar.

T₁=300 K

P₁=1.75 atm

Final temperature of system A= T_A ,

Final temperature of system B= T_B =325K

Final pressure for the systems=T₂

Now, the total initial volume of system-

$\begin{array}{rcl} V_{i} & = & V_{A} + V_{B} \\ = & \frac{n_{A} R T_{1}}{P_{1}} + \frac{n_{B} R T_{1}}{P_{1}} \\ V_{i} & = & \frac{2 n_{A} R T_{1}}{P_{1}} . . . e q (1) \end{array}$

Similarly final volume of system-

$\begin{array}{rcl} V_{f} & = & V_{A} + V_{B} \\ = & \frac{n_{A} R T_{A}}{P_{2}} + \frac{n_{B} R T_{B}}{P_{2}} \\ V_{f} & = & \frac{n_{A} R (T_{A} + T_{B})}{P_{2}} . . . e q (2) \end{array}$

Since, the total volume is constant so as-

$\begin{array}{rcl} V_{i} & = & V_{f} \\ \frac{2 n_{A} R T_{1}}{P_{1}} & = & \frac{n_{A} R (T_{A} + T_{B})}{P_{2}} \\ \frac{2 T_{1}}{P_{1}} & = & \frac{(T_{A} + T_{B})}{P_{2}} . . . (3) \end{array}$