Question

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 498 cm3 of air at atmospheric pressure (1.01×105Pa) and a temperature of 27.0 ?C. At the end of the stroke, the air has been compressed to a volume of 46.4 cm3 and the gauge pressure has increased to 2.80×106 Pa . Compute the final temperature.

Answer #1

Construct two versions of the ideal gas law:

P_{1*}V_{1} = n*R*T_{1}

P_{2*}V_{2} = n*R*T_{2}

from equation 1

n*R = P_{1*}V_{1}/T_{1}

from equation 2

T_{2} = P_{2*}V_{2}/(n*R)

Substitute and simplify:

T_{2} =
T_{1*}P_{2*}V_{2}/(P_{1*}V_{1})

Absolute pressure at state 2 in terms of state 2 gauge pressure
(P_{2}g):

P_{2} = P_{2}g + background pressure

Our background pressure is identical to the pressure at intake
state, thus:

P_{2} = P_{2}g + P_{1}

Thus:

T_{2} = T_{1*}(P_{2}g +
P_{1})*V_{2}/(P_{1*}V_{1})

given data

T_{1}=300.15 K

P_{2}g=2800 kPa; P_{1}=101 kPa

V_{2}=46.4 cm^{3}; V_{1}=498
cm^{3};

T_{2} =300.15 K *(2800 KPa + 101 KPa)*46.4
cm^{3}/(101 KPa*498 cm^{3})

T2 = 803.25 Kelvin

And convert into Celsius:

T2 = 530 Celsius

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