Inquiry into Physics
Inquiry into Physics
8th Edition
ISBN: 9781337515863
Author: Ostdiek
Publisher: Cengage
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Chapter 5, Problem 45Q
To determine

To rank the pressure of each container.

Expert Solution & Answer
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Answer to Problem 45Q

The pressure of the containers is ranked from largest to smallest pressure as,

P1>P3>P5=P6>P4>P2.

Explanation of Solution

Container A:

Given info:

The temperature of the container is 200K.

The volume of the container A is 2000cm3.

The value of universal gas constant R is 8.314 Jmol-1K-1.

Formula used:

Formula to find the pressure of the container.

Using ideal gas equation,

PV=nRT

Here, P is the pressure, V is the volume of the gas, n number of moles, R universal gas constant and T is the temperature the copper container.

Calculation:

Substitute the given values to find the final equilibrium temperature,

Let container contains one mole of helium gas the pressure is,

P1=8.314 Jmol-1K-1×200K2000cm3=1.247atm

Container B:

Given info:

The temperature of the container is 300K.

The volume of the container B is 15000cm3.

The value of universal gas constant R is 8.314 Jmol-1K-1.

Formula used:

Formula to find the pressure of the container.

Using ideal gas equation,

PV=nRT

Here, P is the pressure, V is the volume of the gas, n number of moles, R universal gas constant and T is the temperature the copper container.

Calculation:

Substitute the given values to find the final equilibrium temperature,

Let container contains one mole of helium gas the pressure is,

P2=8.314 Jmol-1K-1×300K15,000cm3=0.1663atm

Container C:

Given info:

The temperature of the container is 400K.

The volume of the container C is 5000cm3.

The value of universal gas constant R is 8.314 Jmol-1K-1.

Formula used:

Formula to find the pressure of the container.

Using ideal gas equation,

PV=nRT

Here, P is the pressure, V is the volume of the gas, n number of moles, R universal gas constant and T is the temperature the copper container.

Calculation:

Substitute the given values to find the final equilibrium temperature,

Let container contains one mole of helium gas the pressure is,

P3=8.314 Jmol-1K-1×300K5,000cm3=0.499atm

Container D:

Given info:

The temperature of the container is 300K.

The volume of the container D is 10,000cm3.

The value of universal gas constant R is 8.314 Jmol-1K-1.

Formula used:

Formula to find the pressure of the container.

Using ideal gas equation,

PV=nRT

Here, P is the pressure, V is the volume of the gas, n number of moles, R universal gas constant and T is the temperature the copper container.

Calculation:

Substitute the given values to find the final equilibrium temperature,

Let container contains one mole of helium gas the pressure is,

P4=8.314 Jmol-1K-1×300K10,000cm3=0.249atm

Container E:

Given info:

The temperature of the container is 200K.

The volume of the container E is 4000cm3.

The value of universal gas constant R is 8.314 Jmol-1K-1.

Formula used:

Formula to find the pressure of the container.

Using ideal gas equation,

PV=nRT

Here, P is the pressure, V is the volume of the gas, n number of moles, R universal gas constant and T is the temperature the copper container.

Calculation:

Substitute the given values to find the final equilibrium temperature,

Let container contains one mole of helium gas the pressure is,

P5=8.314 Jmol-1K-1×200K4,000cm3=0.416atm

Container F:

Given info:

The temperature of the container is 500K.

The volume of the container F is 10,000cm3.

The value of universal gas constant R is 8.314 Jmol-1K-1.

Formula used:

Formula to find the pressure of the container.

Using ideal gas equation,

PV=nRT

Here, P is the pressure, V is the volume of the gas, n number of moles, R universal gas constant and T is the temperature the copper container.

Calculation:

Substitute the given values to find the final equilibrium temperature,

Let container contains one mole of helium gas the pressure is,

P6=8.314 Jmol-1K-1×500K10,000cm3=0.416atm

After calculation the pressure of containers is ranked from largest to smallest pressure as,

P1>P3>P5=P6>P4>P2.

Conclusion:

Thus, the pressure of the containers is ranked from largest to smallest pressure as,

P1>P3>P5=P6>P4>P2.

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Chapter 5 Solutions

Inquiry into Physics

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