Question

A truck tire has a volume of 218 L and is filled with air to 35.0 psi at 295 K. After a drive, the air heats up to 318 K. (b) If the tire volume increases 2.0%, what is the pressure (in psi)?

Answer 1

The pressure (in psi) is 37.1

We are given the following information:

- The initial temperature of the air, $T_{1}=295 \mathrm{~K}$

- The initial pressure, $P_{1}=35.0 \mathrm{psi}$.

- The initial volume, $V_{1}=208 \mathrm{~L}$

- The final temperature, $T_{2}=319 \mathrm{~K}$

- The increase in the volume is $2 \%$, that is, $\Delta V=\frac{2}{100} V_{1}$ where $\Delta V$is the increase in the volume.

The combined gas law states that a fixed amount of an ideal gas obeys the following equation: $\frac{P V}{T}=$ constant, where:

- P is the Pressure of the gas.

- V is the Volume of the gas.

- n is the number of moles of gas.

$R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}=0.0821 \mathrm{~L} \cdot {atm} / \mathrm{mol} \cdot \mathrm{K}$ is the Universal Gas constant.

- T is the absolute temperature of the gas.

The final volume of the air is:

$\begin{aligned}V_{2} &=V_{1}+\Delta V \\&=V_{1}+\frac{2}{100} V_{1} \\&=\frac{102}{100} V_{1}\end{aligned}$

Equating the initial and final state, we have:

$\begin{aligned}\frac{P_{2} V_{2}}{T_{2}} &=\frac{P_{1} V_{1}}{T_{1}} \\\Rightarrow P_{2} &=\frac{V_{1}}{V_{2}} \times \frac{T_{2}}{T_{1}} \times P_{1} \\&=\frac{V_{1}}{102 V_{1} / 100} \times \frac{319 \mathrm{~K}}{295 \mathrm{~K}} \times 35.0 \mathrm{psi} \\& \approx 37.1 \mathrm{psi}\end{aligned}$

The ideal gas law states that a universal constant for an ideal gas is the ratio of the product of pressure and temperature to the product of the number of moles and absolute temperature. The resultant equation is known as the combined gas law if the number of moles in the ideal gas law is set to a constant.

Learn more about combined gas law brainly.com/question/13154969

#SPJ4