Complete Question
A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499cm3 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 46.2cm3 and the gauge pressure has increased to 2.72×106Pa. Compute the final temperature.
Answer:
The final temperature is [tex]T_{final} = 503^o C[/tex]
Explanation:
From the question we are given
The volume of one of the cylinders is [tex]V_1 = 499cm^3[/tex]
The atmospheric pressure is [tex]P_{1} = 1.01 *10^5Pa[/tex]
The temperature is [tex]T = 27^oC = 300K[/tex]
Volume at the end of the stroke is [tex]V_2 = 46.2cm^3 = 319.2 K[/tex]
The increased pressure is [tex]P_2 = 2.72 * 10^6Pa[/tex]
Now to obtain the temperature we are going to apply the ideal gas equation to this question and this is mathematically given as
[tex]PV = nRT[/tex]
Where P is the pressure , V sis the volume T is the temperature
While is the rate constant and n is the number of mole which is constant in this question hence
=> [tex]\frac{PV}{T} = constant[/tex]
[tex]=> \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
Making [tex]T_2[/tex] the subject of the formula we have
[tex]=> T_2 = T_1 (\frac{P_2}{P_1} ) (\frac{V_2}{V_1} )[/tex]
[tex]=> \ \ \ \ \ \ \ \ T_2 = 300(\frac{2.8*10^6Pa}{1.01*10^5Pa} )(\frac{46.2cm^3}{499cm^3} )[/tex]
[tex]= 776K[/tex]
Therefore [tex]T_2 = 776-273 = 503^oC[/tex]
