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Question 9

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An automobile engine contains a vaporized sample of gasoline. The initial pressure in the
cylinder is 42.35 atm at 22.74 Kelvin. After compression, the pressure is increased to
3.52 atm. Calculate the temperature (Kelvin) of the compressed fuel-air mixture.

1 Answer

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Answer and Explanation:

We can use the ideal gas law to solve this problem:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.

First, we need to calculate the initial volume of the gasoline sample. We can assume that the volume of the cylinder remains constant during the compression process, so the initial and final volumes are the same.

We can rearrange the ideal gas law to solve for the number of moles of gas:

n = PV/RT

For the initial state:

n = (42.35 atm) V / (R * 22.74 K)

For the final state:

n = (3.52 atm) V / (R * T2)

where T2 is the final temperature.

Since the number of moles of gas is the same in both states (assuming no gas leaks or additions), we can equate the two expressions for n:

(42.35 atm) V / (R * 22.74 K) = (3.52 atm) V / (R * T2)

We can solve for T2:

T2 = (3.52 atm * 22.74 K) / 42.35 atm = 1.86 K

Therefore, the temperature of the compressed fuel-air mixture is 1.86 Kelvin. Note that this result is unphysical, as it implies that the gasoline has been cooled to an extremely low temperature during the compression process.

User Paul Lassiter
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