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 16, Problem 32SP

One kilomole of ideal gas occupies 22.4 m 3 at 0 °C and 1 atm. (a) What pressure is required to compress 1.00 kmol into a 5.00 m 3 container at 100 °C? (b) If 1.00 kmol was to be sealed in a 5.00 m 3 tank that could withstand a gauge pressure of only 3.00 atm, what would be the maximum temperature of the gas if the tank was not to burst?

(a)

Expert Solution
Check Mark
To determine

The pressure required to compress 1.00 kmol into a 5.00m3 container at 100°C if 1.00 kmol ideal gas occupies 22.4 m3 at 0°C at 1atm pressure.

Answer to Problem 32SP

Solution:

6.12atm

Explanation of Solution

Given data:

The total initial volume of 1.00 kmol gas is 22.4 m3.

The initial temperature of 1.00 kmol gas is 0 °C.

The initial pressure of 1.00 kmol gas is 1atm.

The final volume of 1.00 kmol gas is 5.00m3.

The final temperature of 1.00 kmol gas is 100°C.

Formula used:

Write the ideal gas equation,

PV=nRT

Here, P is the absolute pressure, V is the volume of the gas, n is the number of moles of the gas, and T is the temperature of the gas.

Write the conversion relation for Celsius temperature to Kelvin temperature.

TK=T°C+273.15

Here, TK is the temperature in Kelvin and T°C is the temperature in degree centigrade.

Explanation:

Convert the temperature from °C to K.

For 0 °C,

T1=0+273.15 K=273.15 K

For 100 °C,

T2=100+273.15 K=373.15 K

Recall the expression obtained from the ideal gas law.

PV=nRT

If all other variables are constant then the relation between pressure, volume, and temperature between two states of the gas is written as follows:

P1V1P2V2=T1T2

Here, T1 is the initial temperature, T2 is the final temperature of the gas, V1 is the initial volume, V2 is the final volume of the gas, P1 is the initial pressure, and P2 is the final pressure of the gas.

Rearrange the expression for final pressure.

P2=P1V1T1(T2V2)

Substitute 273.15 K for T1, 1atm for P1, 22.4 m3 for V1, 5.00m3 for V2, and 373.15 K for T2,

P2=(1atm)(22.4 m3)273.15 K(373.15 K5.00m3)=6.12 atm

Conclusion:

Therefore, the final requiredpressure of 1.00 kmol the gas is 6.12atm.

(b)

Expert Solution
Check Mark
To determine

The maximum temperature that the tankwithstands at the pressure of 3atm and volume 5.00m3 when the 1.00 kmol ideal gas which hasinitially occupied 22.4 m3 volume at 0°C temperature, is sealed in the tank.

Answer to Problem 32SP

Solution:

29.26°C

Explanation of Solution

Given data:

The total initial volume of 1.00 kmol gas is 22.4 m3.

The initial temperature of 1.00 kmol gas is 0 °C.

The initial pressure of 1.00 kmol gas is 1atm.

The final volume of 1.00 kmol gas is 5.00m3.

The container can withstands the 3atm pressure only.

Formula used:

Write the ideal gas equation,

PV=nRT

Here, P is the absolute pressure, V is the volume of the gas, n is the number of moles of the gas, and T is the temperature of the gas.

Write the conversion relation for Celsius temperature to Kelvin temperature.

TK=T°C+273.15

Here, TK is the temperature in Kelvin and T°C is the temperature in degree centigrade.

Explanation:

Convert the temperature from °C to K.

For 0 °C,

T1=0+273.15 K=273.15 K

For 100 °C.

T2=100+273.15 K=373.15 K

Calculate the total maximum pressure on the container that withstandsitself.

Pmax=Patm+Pabs

Here, Patm is the atmospheric pressure and Pabs is the absolute pressure.

Substitute 1atm for Patm and 3atm for Pabs

Pmax=1atm+3atm=4atm

Recall the expression obtained from the ideal gas law.

PV=nRT

If all other variables are constant then the relation between pressure, volume, and temperature between twostates of the gas is written as follows:

P1V1PmaxV2=T1Tmax

Here, T1 is the initial temperature, Tmax is the final temperature of the gas, V1 is the initial volume, V2 is the final volume of the gas, P1 is the initial pressure, and Pmax is the final pressure of the gas.

Rearrange the expression for final temperature.

Tmax=PmaxV2P1V1(T1)

Substitute 273.15 K for T1, 1atm for P1, 22.4 m3 for V1, 5.00m3 for V2, and 4atm for Pmax .

Tmax=(4atm)(5.00m3)(1atm)(22.4 m3)(273.15 K)=243.88 K

Convert this temperature into degree Celsius.

For 243.88 K.

T°C=243.884273.1529.26°C

Conclusion:

Therefore, the maximumtemperature of 1.00 kmol the gas is 29.26°C.

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Chapter 16 Solutions

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

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