PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 15, Problem 18P

(a)

To determine

The heat absorbed by the gas.

(a)

Expert Solution
Check Mark

Answer to Problem 18P

The heat absorbed by the gas is 436J_.

Explanation of Solution

Given that the number of moles of the oxygen gas is 1.00mol, the pressure is 1.00atm, the initial temperature is 10.0°C (283.15K), and the final temperature is 25.0°C (298.15K).

Write the expression for the heat absorbed by the gas.

Q=nCpΔT (I)

Here, Q is the heat, n is the number of moles, Cp is the molar specific heat at constant pressure, and ΔT is the change in temperature.

Write the expression for the molar specific heat capacity according to Mayer’s law.

Cp=Cv+R (II)

Here, Cv is the molar specific heat at constant volume, and R is the universal gas constant.

Use equation (II) in (I) and expand the term ΔT

Q=n(Cv+R)(TfTi) (III)

Here, Tf is the final temperature, and Ti is the initial temperature.

Since oxygen is a diatomic gas, the molar specific heat at constant volume is given by,

Cv=52R (IV)

Use equation (IV) in (III).

Q=n(52R+R)(TfTi)=72nR(TfTi) (V)

Conclusion:

Substitute 1.00mol for n, 8.314J/(molK) for R, 298.15K for Tf, and 283.15K for Ti in equation (V) to find Q.

Q=72(1.00mol)[8.314J/(molK)](298.15K283.15K)=436.4J436J

Therefore, the heat absorbed by the gas is 436J_.

(b)

To determine

The change in volume of the gas in the process.

(b)

Expert Solution
Check Mark

Answer to Problem 18P

The change in volume of the gas in the process is 1.23L_.

Explanation of Solution

Given that the number of moles of the oxygen gas is 1.00mol, the pressure is 1.00atm, the initial temperature is 10.0°C (283.15K), and the final temperature is 25.0°C (298.15K).

Write the ideal gas law equation for a gas with constant pressure.

PΔV=nRΔT (VI)

Here, P is the pressure, and ΔV is the change in volume.

Solve equation (VI) for ΔV and expand the term ΔT.

ΔV=nR(TfTi)P (VII)

Conclusion:

Substitute 1.00mol for n, 8.314J/(molK) for R, 298.15K for Tf, 283.15K for Ti, and 1.00atm for P in equation (VII) to find ΔV.

ΔV=(1.00mol)[8.314J/(molK)](298.15K283.15Ki)1.00atm=(1.00mol)[8.314J/(molK)](298.15K283.15Ki)1.00atm×1.013×105Pa1atm×1000L1m3=1.23L

Therefore, the change in volume of the gas in the process is 1.23L_.

(c)

To determine

The work done by the gas during the expansion.

(c)

Expert Solution
Check Mark

Answer to Problem 18P

The work done by the gas during the expansion is 125J_.

Explanation of Solution

Given that the pressure is 1.00atm. It is obtained that the change in volume is 1.23L.

Write the expression for the work done by a gas during expansion.

W=PΔV (VIII)

Here, W is the work done.

Conclusion:

Substitute 1.00atm for P, and 1.23L for ΔV in equation (VIII) to find W.

W=(1.00atm)(1.23L)=(1.00atm×1.013×105Pa1atm)(1.23L×1m31000L)=124.7J125J

Therefore, the work done by the gas during the expansion is 125J_.

(d)

To determine

The change in internal energy of the gas.

(d)

Expert Solution
Check Mark

Answer to Problem 18P

The change in internal energy of the gas is 312J_.

Explanation of Solution

It is obtained that the heat energy absorbed by the gas is 436.4J, and the work done by the gas is 124.7J.

Write the expression for change in internal energy according to first law of thermodynamics.

ΔU=QW (IX)

Here, ΔU is the change in internal energy.

Conclusion:

Substitute 436.4J for Q, and 124.7J for W in equation (IX) to find ΔU.

ΔU=436.4J124.7J=312J

Therefore, the change in internal energy of the gas is 312J_.

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Chapter 15 Solutions

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