a tire contains air ata pressure of 2.8 bar at 10 degree celsius.if the tires volume is unchanged what will the air pressure in it be when the tire warms up to 35 degree celsius if the car is driven.
1 bar = 100000 Pa
2.8 bar =2.8•10^5 Pa
T1= 273+10 =283 K
T2 =273+35 =308 K
p1/T1 =p2/T2
p2=p1•T1/T2 = 2.8•10^5•283/308 =257272.7 Pa
257600
When you cross multiply the original formula you get: p1T2/T1.
Therefore you would get: (2.8 x 10^5)(308)/(283) = p2
3.0
To determine the air pressure in the tire when it warms up from 10°C to 35°C, we can use the ideal gas law. The ideal gas law states that the pressure (P), volume (V), and temperature (T) of a gas are related by the equation PV = nRT, where R is the gas constant and n is the number of moles of gas.
However, since the volume of the tire is unchanged, we can simplify the equation to P₁/T₁ = P₂/T₂, where P₁ and T₁ represent the initial pressure and temperature, and P₂ and T₂ represent the final pressure and temperature.
Given:
Initial pressure (P₁) = 2.8 bar
Initial temperature (T₁) = 10°C = 10 + 273.15 Kelvin (conversion from Celsius to Kelvin)
Final temperature (T₂) = 35°C = 35 + 273.15 Kelvin
Let's plug in the values into the equation and solve for the final pressure:
P₁ / T₁ = P₂ / T₂
2.8 bar / (10 + 273.15 K) = P₂ / (35 + 273.15 K)
Simplifying the equation:
2.8 bar / 283.15 K = P₂ / 308.15 K
Cross-multiplying and solving for P₂:
2.8 bar * 308.15 K = P₂ * 283.15 K
862.42 bar·K = P₂ * 283.15 K
P₂ = 862.42 bar·K / 283.15 K
P₂ ≈ 3.04 bar
Therefore, when the tire warms up to 35°C, the air pressure inside it would be approximately 3.04 bar.