Bundle: College Physics, Loose-Leaf Version, 10th, + WebAssign Printed Access Card for Serway/Vuille's College Physics, 10th Edition, Multi-Term
Bundle: College Physics, Loose-Leaf Version, 10th, + WebAssign Printed Access Card for Serway/Vuille's College Physics, 10th Edition, Multi-Term
10th Edition
ISBN: 9781305367395
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Textbook Question
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Chapter 10, Problem 30P

A 20.0-L tank of carbon dioxide gas (CO2) is at a pressure of 9.50 × 105 Pa and temperature of 19.0°C (a) Calculate the temperature of the gas in Kelvin. (b) Use the ideal gas law to calculate the number of moles of gas in the tank. (c) Use the periodic table to compute the molecular weight of carbon dioxide, expressing it in grams per mole. (d) Obtain the number of grains of carbon dioxide in the tank. (e) A fire breaks out, raising the ambient temperature by 224.0 K while 82.0 g of gas leak out of the tank. Calculate the new temperature and the number of moles of gas remaining in the tank. (f) Using a technique analogous to that in Example 10.6b, find a symbolic expression for the final pressure, neglecting the change in volume of the tank. (g) Calculate the final pressure in the tank as a result of the fire and leakage.

(a)

Expert Solution
Check Mark
To determine
The temperature in Kelvin scale.

Answer to Problem 30P

The temperature in Kelvin scale is 292 K.

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is 19.0οC .

The pressure (P) is 9.50×105Pa

Formula to calculate the temperature in Kelvin scale is,

TK=(T+273.15)K

Substitute 19.0οC for T in the above equation.

TK=(19.0+273.15)K=292K

Conclusion:

The temperature in Kelvin scale is 292 K.

(b)

Expert Solution
Check Mark
To determine
The number of moles of gas.

Answer to Problem 30P

The number of moles of gas is 7.83 mol.

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is 19.0οC .

The pressure (P) is 9.50×105Pa

From Ideal gas equation, formula to calculate the number of moles of gas is,

n=PVRT

  • R is the gas constant.

Substitute 292 K for T, 9.50×105Pa for P, 20.0 L for V and 8.31Jmol1K1 for R in the above equation.

n=(9.50×105Pa)(20.0L)(8.31Jmol1K1)(292K)=(9.50×105Pa)(20.0×103m3)(8.31Jmol1K1)(292K)=7.83mol

Conclusion:

The number of moles of gas is 7.83 mol.

(c)

Expert Solution
Check Mark
To determine
The molecular weight of carbon dioxide (M).

Answer to Problem 30P

The molecular weight of carbon dioxide is 44.0 g/mol.

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is 19.0οC .

The pressure (P) is 9.50×105Pa

The formula of carbon dioxide is CO2 . Therefore, the molecular weight is equal to the molecular weight of carbon and two oxygen atoms.

The molecular weight of carbon is 12.0 g/mol. The molecular weight of oxygen is 16.0 g/mol.

The molecular weight of CO2 is,

M=(12.0g/mol)+2(16.0g/mol)=44.0g/mol

Conclusion:

The molecular weight of carbon dioxide is 44.0 g/mol.

(d)

Expert Solution
Check Mark
To determine
The number of grams of carbon dioxide (m).

Answer to Problem 30P

The number of grams of carbon dioxide is 345 g.

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is 19.0οC .

The pressure (P) is 9.50×105Pa

From (b), n=7.93mol

From (c), M=44.0g/mol .

Formula to calculate the number of grams of carbon dioxide is,

m=nM

Substitute 7.93 mol for n and 44.0 g/mol for M in the above equation to get m.

m=(7.93mol)(44.0g/mol)=345g

Conclusion:

The number of grams of carbon dioxide is 345 g.

(e)

Expert Solution
Check Mark
To determine
The new temperature and number of moles of gas after the leakage.

Answer to Problem 30P

The new temperature is 516 K.

The number of moles of gas after the leakage is 5.98 mol.

Explanation of Solution

Section 1:

To determine: The new temperature.

Answer: The new temperature is 516 K.

Explanation:

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is 19.0οC .

The pressure (P) is 9.50×105Pa

The temperature raises by 224.0 K. 82.0 g of gas leaks out.

From (a), T=292K .

Formula to calculate the new temperature is,

TN=T+(224.0K)

Substitute 292 K for T in the above equation to get m.

TN=(292K)+(224.0K)=516K

The new temperature is 516 K.

Section 2:

To determine: The number of moles of gas after the leakage.

Answer: The number of moles of gas after the leakage is 5.98 mol.

Explanation:

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is 19.0οC .

The pressure (P) is 9.50×105Pa

The temperature raises by 224.0 K. 82.0 g of gas leaks out.

From (d), m=345g

From (c), M=44.0g/mol .

Formula to calculate the number of moles of gas after the leakage is,

n'=m(82.0g)M

Substitute 345 g for m and 44.0 g/mol for M in the above equation.

n'=(345g)(82.0g)44.0g/mol=5.98mol

The number of moles of gas after the leakage is 5.98 mol.

Conclusion:

The new temperature is 516 K.

The number of moles of gas after the leakage is 5.98 mol.

(f)

Expert Solution
Check Mark
To determine
The expression for final pressure.

Answer to Problem 30P

The expression for final pressure is Pi(ViVf)(nfni)(TfTi) .

Explanation of Solution

Given info:

From the ideal gas equation,

PV=nRT

Therefore,

PfVfPiVi=nfRTfniRTi

  • Pi and Pf are the initial and final pressure.
  • Vi and Vf are the initial and final volume.
  • ni and nf are the initial and final number of moles of gas.
  • Ti and Tf are the initial and final temperature.

On Re-arranging the above equation,

Pf=Pi(ViVf)(nfni)(TfTi)

Conclusion:

The expression for final pressure is Pi(ViVf)(nfni)(TfTi) .

(g)

Expert Solution
Check Mark
To determine
The final pressure in the tank as the result of fire and leakage.

Answer to Problem 30P

The final pressure in the tank as the result of fire and leakage is 1.28×106Pa

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is 19.0οC .

The pressure (P) is 9.50×105Pa

The temperature rises by 224.0 K. 82.0 g of gas leaks out.

From (f),

Pf=Pi(ViVf)(nfni)(TfTi) (II)

Since the volume is the same, Vi=Vf .

Substitute 5.98 mol for nf , 7.93 mol for ni , 516 K for Tf , 292 K for Ti , 9.50×105Pa for Pi and Vi for Vf in the Equation (II) to get Pf .

Pf=(9.50×105Pa)i(ViVi)(5.98mol7.93mol)(516K292K)=1.28×106Pa

Conclusion:

The final pressure in the tank as the result of fire and leakage is 1.28×106Pa .

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Chapter 10 Solutions

Bundle: College Physics, Loose-Leaf Version, 10th, + WebAssign Printed Access Card for Serway/Vuille's College Physics, 10th Edition, Multi-Term

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