PHYSICS 1250 PACKAGE >CI<
PHYSICS 1250 PACKAGE >CI<
9th Edition
ISBN: 9781305000988
Author: SERWAY
Publisher: CENGAGE LEARNING (CUSTOM)
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Chapter 21, Problem 21.65AP

A sample of a monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A in Fig. P21.65). It is warmed at constant volume to 3.00 atm (point B). Then it is allowed to expand isothermally to 1.00 atm (point C) and at last compressed isobarically to its original state, (a) Find the number of moles in the sample.

Chapter 21, Problem 21.65AP, A sample of a monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A in Fig.

Find (b) the temperature at point B, (c) the temperature at point C, and (d) the volume at point C. (e) Now consider the processes A → B, B→ C, and C → A. Describe how to carry out each process experimentally, (f) Find Q, W, and ΔEint for each of the processes, (g) For the whole cycle A→ B→ C→ A, find Q, W, and ΔEint.

(a)

Expert Solution
Check Mark
To determine

The number of moles in the sample.

Answer to Problem 21.65AP

The number of moles in the sample is 0.203mol .

Explanation of Solution

Given info: The volume of the monatomic ideal gas is 5.00L at atmospheric pressure and 300K temperature.

The number of moles in the ideal gas equation is,

n=PAVARTA

Here,

PA is the pressure at point A .

VA is the volume at point A .

R is the ideal gas constant.

TA is the temperature at point A .

The value of the ideal gas constant is 8.314J/mol.K .

Substitute 1atm for PA , 5.00L for VA , 300K for TA and 8.314J/mol.K for R in above equation.

n=1atm(101×103Pa1atm)×5.00L(103m31L)8.314J/mol.K×300K=0.203mol

Conclusion:

Therefore, the number of moles in the sample is 0.203mol .

(b)

Expert Solution
Check Mark
To determine

The temperature at point B .

Answer to Problem 21.65AP

The temperature at point B is 900K .

Explanation of Solution

Given info: The volume of the monatomic ideal gas is 5.00L at atmospheric pressure and 300K temperature.

In the process from point A to point B , the volume does not changes.

The expression for the process from point A to point B is,

PATA=PBTB

Here,

PB is the pressure at point B .

TB is the temperature at point B .

Substitute 1atm for PA , 300K for TA , and 3atm for PB in above equation.

1atm300K=3atmTBTB=900K

Conclusion:

Therefore, the temperature at point B is 900K .

(c)

Expert Solution
Check Mark
To determine

The temperature at point C .

Answer to Problem 21.65AP

The temperature at point C is 900K .

Explanation of Solution

Given info: The volume of the monatomic ideal gas is 5.00L at atmospheric pressure and 300K temperature.

In the process from point B to point C , the temperature does not changes because the gas expand isothermally from point B to point C .

So, the temperature at point C will be same as temperature at point at point B which is 900K .

Conclusion:

Therefore, the temperature at point C is 900K .

(d)

Expert Solution
Check Mark
To determine

The volume at point C .

Answer to Problem 21.65AP

The volume at point C is 15.0L .

Explanation of Solution

Given info: The volume of the monatomic ideal gas is 5.00L at atmospheric pressure and 300K temperature.

In the process from point A to point C , the pressure does not changes.

The expression for the process from point A to point C is,

VATA=VCTC

Here,

VC is the volume at point C

TC is the temperature at point C .

Substitute 5.00L for VA , 300K for TA and 900K for TC in above equation.

5.00L300K=VC900KVC=15.0L

Conclusion:

Therefore, the volume at point C is 15.0L .

(e)

Expert Solution
Check Mark
To determine

The experimental methods to carry out the process AB , BC and CA .

Answer to Problem 21.65AP

The experimental method to carry out the process AB is locking the piston, for process BC is releasing the piston to expand at constant temperature, and for process CA is releasing the piston to move freely.

Explanation of Solution

Given info: The volume of the monatomic ideal gas is 5.00L at atmospheric pressure and 300K temperature.

In the process from point A to point B :

The volume does not change. The temperature varies from 300K to 900K between these points. This process can be obtained by locking the piston so that the volume o gas does not changes and pressure increase from 1atm to 3atm .

In the process from point B to point C :

The temperature does not change. The pressure varies from 3atm to 1atm between these points. This process can be obtained by releasing the piston to expand at constant temperature.

In the process from point C to point A :

The pressure does not change. The temperature varies from 900K to 300K between these points. This process can be obtained by releasing the piston to move freely so that after volume varies from 15L to 5L .

Conclusion:

Therefore, the experimental method to carry out the process AB is locking the piston, for process BC is releasing the piston to expand at constant temperature, and for process CA is releasing the piston to move freely.

(f)

Expert Solution
Check Mark
To determine

The heat Q , the work done W and change in internal energy ΔEint for each of the process.

Answer to Problem 21.65AP

The heat Q , the work done W and change in internal energy ΔEint for process AB are 1.52kJ , 0, 1.52kJ respectively and for process BC are 1.67kJ , 1.67kJ ,0 respectively and for process CA are 2.53kJ , 1.01kJ , 1.52kJ .

Explanation of Solution

Given info: The volume of the monatomic ideal gas is 5.00L at atmospheric pressure and 300K temperature.

For the process from point A to point B :

The volume of gas does not change.

The work done is,

W1=0

The change in internal energy is equal to the heat.

The expression for the change in internal energy is,

ΔEint1=Q1=nCv(TBTA)

Here,

n is the number of moles.

Cv is the coefficient of heat transfer at constant volume.

Substitute 32R for Cv , 0.203 for n , 900K for TB and 300K for TA in above equation.

ΔEint1=0.203×32R×(900K300K)=182.7R

Substitute 8.314J/mol.K for R in above equation.

ΔEint1=182.7×8.314J/mol.K=1518.96J(1kJ1000J)=1.52kJ

Thus, change in internal energy in process from point A to point B is 1.52kJ .

For the process from point B to point C :

The temperature does not change.

The change in internal energy is,

ΔEint2=0

The expression of the work done is,

W=Q2=nRTBln(VCVB)

Substitute 8.314J/mol.K for R , 0.203 for n , 900K for TB , 15L for VC and 5L for VB in above equation.

W2=0.203×8.314J/mol.K×900K×ln(15L5L)=1668.76J(1kJ1000J)=1.67kJ

Thus the change in internal energy in process from point B to point C is 1.67kJ .

For the process from point C to point A :

The formula of work done is,

W3=PA(VCVA)

Substitute 1atm for PA , 15L for VC and 5L for VA in above equation.

W3=1atm(101×103pa1atm)×(15L5L)(103m31L)=1010J(1kJ1000J)=1.01kJ

Thus, the work done for the point C to point A is 1.01kJ .

The formula for the change in kinetic energy is,

ΔEint3=32nR(TATC)

Substitute 8.314J/mol.K for R , 0.203 for n , 900K for TC and 300K for TA in above equation.

ΔEint3=32×0.203×8.314J/mol.K×(300K900K)=1518.97J(1kJ1000J)=1.52kJ

The heat obtain in this process is,

Q3=ΔEint3W3=1.51kJ1.01kJ=2.53kJ

Conclusion:

Therefore, the heat Q , the work done W and change in internal energy ΔEint for process AB are 1.52kJ , 0, 1.52kJ respectively and for process BC are 1.67kJ , 1.67kJ ,0 respectively and for process CA are 2.53kJ , 1.01kJ , 1.52kJ .

(g)

Expert Solution
Check Mark
To determine

The heat Q , work done W and ΔEint for the whole cycle ABCA .

Answer to Problem 21.65AP

For the whole cycle ABCA , the heat Q , work done W and ΔEint are 0.656kJ , 0.656kJ and 0 respectively.

Explanation of Solution

Given info: The volume of the monatomic ideal gas is 5.00L at atmospheric pressure and 300K temperature.

The expression for the heat in complete cycle is,

Q=Q1+Q2+Q3

Substitute 1.51kJ for Q1 , 1.66kJ for Q2 and 2.52kJ for Q3 in above equation.

Q=1.51kJ+1.66kJ2.52kJ=0.65kJ

Thus, the heat in cycle is 0.65kJ .

The expression for the work done in complete cycle is,

W=W1+W2+W3

Substitute 0 for W1 , 1.66kJ for W2 and 1.01kJ for W3 in above equation.

W=01.66kJ+1.01kJ=0.65kJ

Thus, the total work done is 0.65kJ .

As the process is cyclic, the change in internal energy will be zero.

ΔEint=0

Conclusion:

Therefore, For the whole cycle ABCA , the heat Q , work done W and ΔEint are 0.656kJ , 0.656kJ and 0 respectively.

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Chapter 21 Solutions

PHYSICS 1250 PACKAGE >CI<

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