Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Chapter 20, Problem 51SP

(a)

To determine

The temperature of the gas at B if the temperature at A is 400°C in Fig. 20-6.

(a)

Expert Solution
Check Mark

Answer to Problem 51SP

Solution:

2.0×103 K

Explanation of Solution

Given data:

Refer to Fig. 20-6.

The temperature at A is 400°C.

The mass of gas enclosed in the cylinder is 4 g.

The gas follows the process A to B in the thermodynamic cycle shown in Fig. 20-6.

Formula used:

The gas equation for a process is expressed as

P1V1T1=P2V2T2

Here, P1 represent the initial pressure, V1 represents the initial volume, T1 represents the initial temperature, P2 represent the final pressure, V2 represents the final volume, and T2 represents the final temperature.

The formula for the conversion of the initial temperature of a gas from the Celsius scale to the Kelvin scale is

T1=[t1+273] K

Here, t1 is the initial temperature of the gas in Celsius.

Explanation:

Draw the thermodynamic cycle diagram given in Fig- 20.6:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 20, Problem 51SP , additional homework tip  1

Recall the expression for the conversion of temperature at A from Celsius to Kelvin:

TA=[tA+273] K

Here, TA is the temperature at point A in Kelvin and tA is the temperature at point A in Celsius.

Substitute 400°C for tA

TA=[400°C+273] K=673 K

Refer to the diagram and write the values of pressure and volume at points A and B, respectively,

PA=6 atmPB=4 atm

And

VA=2 litresVB=9 litres

Here, PA pressure at the point A, VA represent the volume at point A, PB represents the pressure at the point B, and VB represent the volume at point B.

Recall the gas equation between points A and B:

PAVATA=PBVBTB

Here, TB is the temperature at point B in Kelvin.

Substitute 4 atm for PB, 9 litres for VB, 6 atm for PA, 673 K for TA, and 2 litres for VA

(6 atm)(2 litres)673 K=(4 atm)(9 litres)TBTB=(4 atm)(9 litres)(6 atm)(2 litres)(673 K)2.0×103 K

Conclusion:

The temperature at point B is 2.0×103 K.

(b)

To determine

The value of cv for the 4 g of gas enclosed in the cylinder if the portion from A to B receives 2.20 kcal of heat, in Fig. 20-6.

(b)

Expert Solution
Check Mark

Answer to Problem 51SP

Solution:

0.25 cal/(g°C)

Explanation of Solution

Given data:

Refer to Fig. 20-6.

The temperature at A is 400°C.

The heat received by the gas from A to B is 2.20 kcal.

The mass of the gas enclosed in the cylinder is 4 g.

Formula used:

The area of a trapezium is calculated by the formula:

A=12(a+b)(h)

Here, A is the area of the trapezium, a and b are the lengths of the parallel sides of the trapezium, a+b represents the sum of the parallel sides, and h is the perpendicular distance between the parallel sides of the trapezium.

The work done in a thermodynamic process is given by the area under the line representing the process in the pressure–volume diagram:

ΔW=APVD

Here, APVD is the area under the line representing the process in the pressure–volume diagram and ΔW is the work done in the process.

The first law of thermodynamics for a process is written as

ΔQ=ΔU+ΔW

Here, ΔU is the change in the internal energy during the process and ΔQ is the heat supplied to the system during the process.

The formula for change in internal energy is

ΔU=mcvΔT

Here, ΔT is the change in temperature during the process, cv is the specific heat at constant volume, and m is the mass of the body.

The formula for conversion of temperature of gas from Kelvin scale to Celsius scale is

t=[T273] °C

Here, t is the temperature in Kelvin and T is the temperature in degree Celsius.

Explanation:

Draw the thermodynamic cycle diagram given in Fig- 20.6:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 20, Problem 51SP , additional homework tip  2

Understand that the work done in the thermodynamic process AB is equal to the area of the pressure–volume diagram under the line AB.

Draw the thermodynamic cycle diagram showing the area under the line AB:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 20, Problem 51SP , additional homework tip  3

Here, the points E and F are as shown in the figure and the work done during the process AB is represented by the area under the line AB, which is equal to the area of the trapezium ABEF.

Refer to the figure and write the values of the lengths of sides AF, BE, and EF:

AF=6 atm0 atm=6 atm

BE=4 atm0 atm=4 atm

EF=9 litres2 litres=7 litres

Recall the expression for the area of trapezium ABEF to calculate the area under the line AB in order to calculate the work done in the process AB:

A1=12(AF+BE)(EF)

Here, A1 is the area of the trapezium ABEF.

Substitute 6 atm for AF, 4 atm for BE, and 7 litres for EF

A1=12(6 atm+4 atm)(7 litres)=12(10 atm)(7 litres)=12[(10 atm)(1.013×105 Pa1 atm)][(7 litres)(103 m31 litre)]=3545.5 J

Recall the expression for the net-work done in a thermodynamic process in terms of the area of the pressure–volume diagram:

ΔW=A1

Substitute 3545.5 J for A1

ΔW=3545.5 J=3545.5 J(0.001 kJ1 J)=3.545 kJ

Recall the expression for the first law of thermodynamics for the process AB:

ΔQ=ΔU+ΔW

According to the problem, the heat supplied from A to B is 2.20 kcal.

Substitute 2.20 kcal for ΔQ and 3.545 kJ for ΔW

2.20 kcal=ΔU+3.545 kJΔU=2.20 kcal3.545 kJ=2.20 kcal(1000 cal1 kcal)(3.545 kJ)(14.186 kcal1kJ)(1000 cal1 kcal)=2200 cal846.9 cal

Further solve as

ΔU=1353.1 cal

Calculate the temperature at point B in Celsius:

tB=[TB273]°C

Here, tB is the temperature at point B in Celsius.

Substitute 2.0×103 K for TB

tB=[2.0×103 K273]°C=1.727×103°C=1727°C

Calculate the change in temperature from A to B:

ΔT=tBtA

Substitute 1727°C for tB and 400°C for tA

ΔT=1727°C400°C=1327°C

Recall the formula for change in internal energy:

ΔU=mcvΔT

Substitute 1353.1 cal for ΔU, 1327°C for ΔT, and 4 g for m

1353.1 cal=(4 g)cv(1327°C)cv=1353.1 cal(4 g)(1327°C)=1353.1 cal(4 g)(1327°C)0.25 cal/(g°C)

Conclusion:

The value of cv for the gas is 0.25 cal/(g°C).

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Chapter 20 Solutions

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)

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