Um motorista calibra os pneus de seu carro com uma pressão de 30 libras/pol^2 a uma temperatura de 27 ºC. Após uma viagem, a temperatura deles subiu para 47 ºC. Desprezando-se a variação de volume dos pneus e sabendo-se que 10% da massa de ar contida em um dos pneus escapou pela válvula durante a viagem, a pressão do ar neste pneu, ao término desta viagem, em libras/pol^2, é de aproximadamente

Let's break down this problem step by step. We have a motorist who calibrates the tires of his car with a pressure of 30 lb/in² at a temperature of 27°C. After a trip, the temperature rises to 47°C. We need to find the pressure of the air in one of the tires at the end of the trip, considering that 10% of the air mass escaped through the valve during the trip.

First, let's use Charles' Law, which states that, at constant volume, the pressure of a gas is directly proportional to the temperature. Mathematically, this is represented as:

$$P_1 / T_1 = P_2 / T_2$$

where $P_1$ and $T_1$ are the initial pressure and temperature, and $P_2$ and $T_2$ are the final pressure and temperature.

We know that the initial pressure is 30 lb/in², and the initial temperature is 27°C = 300 K (we convert the temperature to Kelvin). The final temperature is 47°C = 320 K. We can set up the equation as:

$$30 / 300 = P_2 / 320$$

Solving for $P_2$, we get:

$$P_2 = (30 * 320) / 300 = 32$$

However, we need to consider that 10% of the air mass escaped during the trip. This means that the final pressure will be lower than the calculated value. Let's assume the initial air mass is $m_0$, and the air mass that escaped is 0.1$m_0$. The final air mass is then $m_f = m_0 - 0.1m_0 = 0.9m_0$.

Using the ideal gas law, we can write:

$$P_1V = n_1RT_1$$

and

$$P_2V = n_2RT_2$$

where $n_1$ and $n_2$ are the initial and final number of moles of gas, respectively.

Since the volume is constant, we can divide both equations by $V$ and rearrange to get:

$$P_1 / T_1 = (n_1R) / V$$

and

$$P_2 / T_2 = (n_2R) / V$$

Substituting the values, we get:

$$(30) / (300) = (n_1R) / V$$

and

$$(P_2) / (320) = (0.9n_1R) / V$$

Equating both equations, we can solve for $P_2$:

$$P_2 = (30 * 320) / (300 * 0.9) = 29$$

Therefore, the correct answer is alternative C) 29 lb/in².

Remember that Charles' Law is a fundamental principle in thermodynamics, and it's essential to understand how to apply it in problems like this one. Also, pay attention to the details, such as the fact that 10% of the air mass escaped, which affects the final pressure.