In an internal combustion engine, air at atmostpheric

Physics

1 answer:

notsponge [240]3 years ago

8 0

Answer:

1302 K or 1029 C

Explanation:

Air at atmospheric pressure has pressure of 1 atm

20 C = 20 + 273 = 293 K

Assume ideal gas, according to the ideal gas law:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Where P1, V1 and T1 are the pressure, volume and temperature of the gas before the compression and P2, V2 and T2 are the pressure, volume and temperature of the gas after the compression

$T_2 = T_1\frac{P_2V_2}{P_1V_1}$

Since the gas is compressed to 1/9 of its original volume, V2/V1 = 1/9:

$T_2 = 293\frac{40}{9} = 1302 K$ or 1029 C

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