A diesel engine's piston compresses 16 cm3 of fuel-air mixture into 1 cm3. The pressure changes from 1 atmosphere to 48 atmospheres. If the initial temperature of the gas was 305 K, what was the final temperature? Note: As long as the units for pressure and volume are the same on both sides of the equation, they will cancel. Temperature, however, must be in units of kelvin. Be sure to use the proper number of signifi A diesel engine's piston compresses 16 cm3 of fuel-air mixture into 1 cm3. The pressure changes from 1 atmosphere to 48 atmospheres. If the initial temperature of the gas was 305 K, what was the final temperature? Note: As long as the units for pressure and volume are the same on both sides of the equation, they will cancel. Temperature, however, must be in units of kelvin. Be sure to use the proper number of significant figures. 915 K 1.09 x 10-3 K 900 K 100 Kcant figures. 915 K 1.09 x 10-3 K 900 K 100 K

Question

Chemistry

Briley Gibson

1 August, 20:14

A diesel engine's piston compresses 16 cm3 of fuel-air mixture into 1 cm3. The pressure changes from 1 atmosphere to 48 atmospheres. If the initial temperature of the gas was 305 K, what was the final temperature? Note: As long as the units for pressure and volume are the same on both sides of the equation, they will cancel. Temperature, however, must be in units of kelvin. Be sure to use the proper number of signifi A diesel engine's piston compresses 16 cm3 of fuel-air mixture into 1 cm3. The pressure changes from 1 atmosphere to 48 atmospheres. If the initial temperature of the gas was 305 K, what was the final temperature? Note: As long as the units for pressure and volume are the same on both sides of the equation, they will cancel. Temperature, however, must be in units of kelvin. Be sure to use the proper number of significant figures. 915 K 1.09 x 10-3 K 900 K 100 Kcant figures. 915 K 1.09 x 10-3 K 900 K 100 K

+4

Answers (1)

Know the Answer?

Jordin Weaver · Answer 1

Answer is:

the final temperature was 915 K.

p₁ = 1 atm.

T₁ = 305 K.

V₁ = 16 cm³

p₂ = 48 atm.

V₂ = 1 cm³.

T₂ = ?

Ideal law gas:

p₁V₁ / T₁ = p₂V₂ / T₂.

1 atm · 16 cm³ / 305 K = 48 atm · 1 cm³ / T₂.

T₂ = 305 K · 48 atm·cm³ : 16 atm·cm³.

T₂ = 915 K.

p - pressure. of fuel air

T - temperature of fuel air.

V - volume of fuel air.