Question 13Consider a car at a chilly temperature of 44 °F. If the pressure is 101 kPa, what is the density of air in the car? How would increasing the temperature to 77°F change the density of air in the car?1 point1124 g/m^3, decrease by 98 g/m^31124 g/m^3, increase by 98 g/m^31257 g/m^3, decrease by 77 g/m^31257 g/m^3, increase by 77 g/m^3

To answer this question, we need to use the ideal gas law equation, which is PV = nRT. However, we need it in a form that allows us to solve for density (ρ), so we'll use the equation ρ = PM/RT, where P is pressure, M is molar mass, R is the ideal gas constant, and T is temperature.

First, we need to convert all our units to be consistent. The molar mass of air is approximately 28.97 g/mol. The ideal gas constant R is 8.314 J/(mol·K), but we need it in units of L·kPa/(K·mol), so R = 8.314 kPa·L/(mol·K). The pressure P is given as 101 kPa. The temperature needs to be in Kelvin, so we convert 44°F to Kelvin by using the formula K = (°F + 459.67) * 5/9, which gives us approximately 279.82 K.

Now we can plug these values into our equation:

ρ = (101 kPa * 28.97 g/mol) / (8.314 kPa·L/(mol·K) * 279.82 K)

Solving this gives us a density of approximately 1.184 kg/m^3 or 1184 g/m^3.

Next, we need to find out how the density changes when the temperature increases to 77°F. We convert 77°F to Kelvin, which is approximately 298.15 K. Plugging this into our equation gives us:

ρ = (101 kPa * 28.97 g/mol) / (8.314 kPa·L/(mol·K) * 298.15 K)

Solving this gives us a density of approximately 1.127 kg/m^3 or 1127 g/m^3.

So, the density of the air in the car decreases by approximately 57 g/m^3 when the temperature increases from 44°F to 77°F. Therefore, none of the provided options are correct.