a 64.0L gas tank is filled at 6.00C. the next day the temperature gets up to 27.00C how much gas (octance) voerflows?
To determine the amount of gas (octane) that overflows from the gas tank due to the change in temperature, you need to use the ideal gas law equation and apply the concept of thermal expansion.
The ideal gas law equation is:
PV = nRT
where:
P = pressure of the gas (constant)
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature of the gas (in Kelvin)
Since we have the volume (V) of the gas tank and the initial and final temperatures (6.00 °C and 27.00 °C, respectively), we can find the final volume of the gas by converting the temperatures to Kelvin and applying the formula for thermal expansion.
First, let's convert the temperatures from Celsius to Kelvin. Remember that Kelvin is an absolute temperature scale, where 0 K is absolute zero and there are no negative values.
Initial temperature (T1) = 6.00 °C + 273.15 = 279.15 K
Final temperature (T2) = 27.00 °C + 273.15 = 300.15 K
Now, let's calculate the change in volume (ΔV) due to the temperature increase by using the formula for thermal expansion:
ΔV = V * β * ΔT
where:
ΔV = change in volume
V = initial volume of the gas tank (64.0 L in this case)
β = coefficient of thermal expansion (depends on the substance)
ΔT = change in temperature (T2 - T1)
The coefficient of thermal expansion for gasoline is not readily available, but we can assume it to be approximately 0.001 (1/1000, which is a general estimate for liquids).
Using these values, we can calculate the change in volume (ΔV):
ΔV = 64.0 L * 0.001 * (300.15 K - 279.15 K)
ΔV = 64.0 L * 0.001 * 21 K
ΔV ≈ 1.344 L
Therefore, approximately 1.344 liters of gas (octane) will overflow from the gas tank when the temperature rises from 6.00 °C to 27.00 °C.