3. 0.03 kg of air are adiabatically compressed in the cylinder of a diesel engine at the start of compression the volume of the cylinder filled with air is 0.03m^3 and the pressure at the end of compression is 3 .64 MPa. What will be the temperature of the air at the end of compres- sion if its volume decreases by a factor of 13.5 ?

Question

3. 0.03 kg of air are adiabatically compressed in the cylinder of a diesel engine at the start of compression the volume of the cylinder filled with air is 0.03m^3 and the pressure at the end of compression is 3 .64 MPa. What will be the temperature of the air at the end of compres- sion if its volume decreases by a factor of 13.5 ?

Answer 1

To solve this problem, we can use the ideal gas law and the adiabatic compression equation.Given information:- Initial volume of the cylinder filled with air, V1 = 0.03 m^3- Final pressure at the end of compression, P2 = 3.64 MPa- Volume decreases by a factor of 13.5, so the final volume, V2 = V1 / 13.5We need to find the final temperature of the air at the end of compression, T2.The ideal gas law states that the product of pressure and volume is proportional to the temperature, which can be written as:P1V1/T1 = P2V2/T2Since the process is adiabatic, there is no heat transfer, and the specific heat ratio (γ) remains constant. The adiabatic compression equation can be written as:P1V1^(γ) = P2V2^(γ)Substituting the given values and solving for T2, we get:T2 = (P2 * V2^(γ)) / (P1 * V1^(γ))T2 = (3.64 MPa * (0.03 m^3 / 13.5)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa00222 m)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T2 = (3.64 MPa * (0.00222 m^3)^(γ)) / (1 MPa * (0.03 m^3)^(γ))T

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