During the compression stroke of a certain gasoline engine, the pressure increases from 1.00 atm to 20.5 atm. The process is adiabatic and the air–fuel mixture behaves as a diatomic ideal gas.
(a) By what factor does the volume change?
Vfinal = Vinitial
(b) By what factor does the temperature change?
Tfinal = Tinitial
Assume the compression starts with 0.016 mole of gas at 26.5°C.
(c) Find the value of Q that characterizes the process.
J
(d) Find the value of ΔEint that characterizes the process.
J
(e) Find the value of W that characterizes the process.
J
Adiabatic index
γ=(i+2)/i=(5+2)/5= 1.4.
(a) p₁(V₁)^γ=p₂(V₂)^γ
V₁/V₂ =( p₂/p₁)^(1/ γ)=20.5^(1/1.4)=
=20.5^(0.714)=8.65.
(b) (p₂/p₁)^(γ-1) = (T₂/T₁)^γ
(T₂/T₁) =((p₂/p₁)^[(γ-1/γ)]=
=20.5^(0.2857)=2.37.
(c) An adiabatic process is a process occurring without exchange of heat of a system with its environment =>
Q=0.
ν =0.016 mol
T₁=273+26.5 = 299.5 K,
T₂/T₁=2.37,
T₂=2.37•T₁=2.37•299.5=710 K.
(d,e) ΔE=- W
W= ν•{R/(γ-1)}(T₁-T₂) =
=0.016•8.31(299.5-710)/(1.4-1) =
= - 136.45 J
ΔE=136.45 J
To solve this problem, we can use the adiabatic compression equation for an ideal gas:
P₁V₁^γ = P₂V₂^γ
where P₁ and V₁ are the initial pressure and volume, P₂ and V₂ are the final pressure and volume, and γ is the heat capacity ratio of the gas.
Step 1: Find the initial and final volumes
(a) By rearranging the equation, we can find the factor by which the volume changes:
V₂ / V₁ = (P₁ / P₂)^(1/γ)
V₂ / V₁ = (1.00 atm / 20.5 atm)^(1/γ)
To find γ, we need to know the specific gas in the mixture. Since it is not provided, we cannot calculate the exact value.
(b) Similarly, we can calculate the factor by which the temperature changes using the adiabatic relations:
T₂ / T₁ = (V₁ / V₂)^(γ-1)
T₂ / T₁ = (V₁ / V₂)^(γ-1)
To solve this, we need to know γ as well.
Step 2: Calculate the value of Q
(c) Since the process is adiabatic, it means no heat is exchanged with the surroundings. Therefore, Q = 0 J.
Step 3: Calculate the change in internal energy
(d) ΔEint = Q + W, where ΔEint is the change in internal energy, Q is heat, and W is work.
Since Q = 0 J (from step 2) and the process is adiabatic, ΔEint = W.
(e) To calculate W, we use the equation:
W = P₁V₁ - P₂V₂ / (γ - 1)
To calculate W, we need the value of γ.
Therefore, without knowing the specific gas and its heat capacity ratio (γ), we cannot determine the exact answers to parts (a), (b), (d), and (e) of the question.
(a) To find the factor by which the volume changes during the adiabatic compression, you can use the adiabatic equation for ideal gases:
(P1 * V1^γ) = (P2 * V2^γ)
Where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and γ is the adiabatic index for the gas.
In this case, P1 = 1.00 atm, P2 = 20.5 atm. Since the air-fuel mixture behaves as a diatomic ideal gas, γ = 7/5.
By rearranging the equation, you can solve for the factor by which the volume changes:
V2 / V1 = (P1 / P2)^(1/γ)
V2 / V1 = (1.00 atm / 20.5 atm)^(1/7/5)
V2 / V1 ≈ 0.292
Therefore, the volume changes by a factor of approximately 0.292.
(b) To find the factor by which the temperature changes during the adiabatic compression, you can use the ideal gas law:
P * V = n * R * T
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
In this case, we assume the number of moles of gas is constant at 0.016 mole, the gas constant R is 8.314 J/(mol·K), and the initial temperature T1 is 26.5°C, which needs to be converted to Kelvin.
T1 = 26.5°C + 273.15 = 299.65 K
By rearranging the ideal gas law equation:
T2 / T1 = (P2 / P1) * (V1 / V2)
T2 = T1 * (P2 / P1) * (V1 / V2)
T2 = 299.65 K * (20.5 atm / 1.00 atm) * (1 / 0.292)
T2 ≈ 2175 K
Therefore, the temperature changes by a factor of approximately 2175 K / 299.65 K, or approximately 7.25.
(c) The value of Q that characterizes the process is the heat transferred during the adiabatic compression. Since the process is adiabatic, it means no heat is transferred, so Q = 0 J.
(d) The value of ΔEint that characterizes the process is the change in internal energy. For an adiabatic process, the change in internal energy is given by the equation:
ΔEint = Q - W
Since Q = 0 J (as explained in part c) and the work, W, is given by the equation:
W = -ΔP * ΔV
Where ΔP is the change in pressure and ΔV is the change in volume.
In this case, ΔP = 20.5 atm - 1.00 atm = 19.5 atm, and ΔV = V2 - V1, since we have already found the factor by which the volume changes in part a.
ΔV = V1 * (1 - 0.292)
Plugging in the values:
ΔEint = 0 J - (-19.5 atm * ΔV)
ΔEint = 19.5 atm * V1 * 0.708
ΔEint ≈ 13.9 atm * V1
Therefore, the value of ΔEint that characterizes the process is approximately 13.9 atm * V1.
(e) The value of W that characterizes the process is the work done on the system during the adiabatic compression. As mentioned before, the work is given by:
W = -ΔP * ΔV
Plugging in the values of ΔP and ΔV:
W = -19.5 atm * V1 * 0.708
W ≈ -13.8 atm * V1
Therefore, the value of W that characterizes the process is approximately -13.8 atm * V1. The negative sign indicates that work is done on the system.