Chapter U3.61, Problem 4E

(a)

Interpretation Introduction

Interpretation: Whether the pressure of tire increases or decreases need to be explained.

Concept Introduction: The combined gas law is combination of three gas laws that is Boyle’s law, Charles’ law and Gay-Lussac’s law. According to this law the ratio of temperature to the product of pressure and volume is a constant.

Explanation of Solution

The initial pressure and temperature is 1.0 atm and 21 ^oC respectively. The final temperature is 55 ^oC. The volume and number of moles of gas in the tire is constant.

According to Gay-Lussac’s law, at constant volume, pressure and temperature of the gas are directly proportional to each other.

Here, temperature changes from 21 ^oC to 55 ^oC thus, it increases and pressure also increases.

(b)

Interpretation Introduction

Interpretation: The equation used to calculate the new pressure of the tire needs to be determined.

Concept Introduction: The combined gas law is combination of three gas laws that is Boyle’s law, Charles’ law and Gay-Lussac’s law. According to this law the ratio of temperature to the product of pressure and volume is a constant.

Explanation of Solution

According to Gay-Lussac’s law, at constant volume, pressure and temperature of the gas are directly proportional to each other This is mathematically represented as follows:

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Here, T₁ is initial volume, T₂ is final volume, P₁ is initial pressure and P₂ is final pressure.

Thus, the following equation can be used to determine the new pressure:

  $P_2=\frac{P_1\times T_2}{T_1}$

(c)

Interpretation Introduction

Interpretation: The pressure of the tire at 55 ^oC needs to be determined.

Concept Introduction: According to Gay-Lussac’s law, at constant volume, pressure and temperature of the gas are directly proportional to each other This is mathematically represented as follows:

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Here, T₁ is initial volume, T₂ is final volume, P₁ is initial pressure and P₂ is final pressure.

Explanation of Solution

The initial pressure and temperature is 1.0 atm and 21 ^oC respectively. The final temperature is 55 ^oC.

The given initial and final temperature can be converted into Kelvin as follows:

  $T_1=21+273.15=294.15\text{ K}$

And,

  $T_2=55+273.15=328.15\text{ K}$

The final pressure can be calculated as follows:

  $P_2=\frac{P_1\times T_2}{T_1}$

Putting the values,

  $P_2=\frac{\left(1\text{ atm}\right)\left(328.15\text{ K}\right)}{\left(294.15\text{ K}\right)}=1.12\text{ atm}$

Therefore, the new pressure is 1.12 atm.