One cylinder in the diesel engine of a truck has an initial volume of 600 cm 3.

Question

One cylinder in the diesel engine of a truck has an initial volume of 600 cm 3. Air is admitted to the cylinder at 300 C and a pressure of 1.0 atm. The piston rod then does 400 J of work to rapidly compress the air. What are its final temperature and volume?

Explanation / Answer

i entered sample problem here plese substitute correct values instead of original values like air is 35->300,,,the piston rod 500j->400j etc First law of thermodynamics: Q = ?U - W Solve for ?U: ?U = Q - W Because it is adiabatic, Q = 0. Because work was done on the system, the given value of work is actually a negative number. We will later plug in a negative value of W. Thus: ?U = -W Relation of ?U to change in temperature (assuming air to be calorically perfect): ?U = n*c_v*?T Solve for ?T: ?T = ?U/(n*c_v) To get c_v, the molar specific heat capacity, use the following (knowing value of R and adiabatic index): c_v = R/(k - 1) ?T = ?U*(k - 1)/(n*R) So, our final temperature: T2 = T1 + ?T T2 = T1 + ?U*(k - 1)/(n*R) T2 = T1 - W*(k - 1)/(n*R) And to achieve n*R, use ideal gas law: P1*V1 = n*R*T1 Solve for n*R: n*R = P1*V1/T1 Thus: T2 = T1*(1 - W*(k - 1)/(P1*V1)) To find corresponding final volume, ASSUME our adiabatic process is an adiabatic and reversible process (it isn't necessarily adiabatic and reversible, but we don't have the information to prove otherwise). This means that the formula for work is: W = (P2*V2 - P1*V1)/(1 - k) And the relation of state 1 and state 2: P2*V2^k = P1*V1^k Solve for P2: P2 = P1*(V1/V2)^k Thus: W = (P1*(V1/V2)^k*V2 - P1*V1)/(1 - k) Factor: W = P1*(V2*(V1/V2)^k - V1)/(1 - k) Simplify the term: V2*(V1/V2)^k V2*(V1/V2)^k V1^k*V2/V2^k V1*V1^(k-1)*V2^(1 - k) V1*(V2/V1)^(1 - k) Substitute: W = P1*(V1*V2^(1 - k) - V1)/(1 - k) Factor: W = P1*V1*((V2/V1)^(1 - k) - 1)/(1 - k) Solve for V2: W*(1 - k)/(P1*V1) = (V2/V1)^(1 - k) 1 + W*(1 - k)/(P1*V1) = (V2/V1)^(1 - k) V2/V1 = (1 + W*(1 - k)/(P1*V1))^(1/(1 - k)) V2 = V1*(1 + W*(1 - k)/(P1*V1))^(1/(1 - k)) Summary of equations providing results: T2 = T1*(1 - W*(k - 1)/(P1*V1)) V2 = V1*(1 + W*(1 - k)/(P1*V1))^(1/(1 - k)) Data (notice unit conversions?): W:= -500 J; P1:=101325 Pa; V1:=0.0006 m^3; T1:=308.15 K; k:=1.4 (standard value for air); Results: T2 = 1322 Kelvin V2 = 15.74 cm^3

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