Question
A heat engine takes 0.400 mol of an ideal diatomic gas around the cycle shown in the pV diagram of the figure (Figure 1) . Process 1->2 is at constant volume, process 2->3 is adiabatic, and process 3->1 is at a constant pressure of 1.00 atm. The value of gamma for this gas is 1.40.
Find the pressure at points 1, 2, and 3
Find the volume at points 1, 2, and 3.
Explanation / Answer
1. p1 and p3 are given in the figure: p1 = p3 = 1atm To calculate p2 use ideal gas law p·V = n·R·T the the amount n stays throughout the whole process p·V/T = n·R = constant Hence p1·V1/T1 = p2·V2/T2 because V1= V2 p1/T1 = p2/T2 => p2 = p1·T2/T1 = 1atm · 600K/300K = 2atm 2. use ideal gas law to calculate the volume: V = n·R·T/p V1 = n·R·T1/p1 = 0.35mol · 0.0820574587atmL/molK · 300K / 1atm = 8.616L V2 = V1 = 8.616L V3 = n·R·T3/p3 = 0.35mol · 0.0820574587atmL/molK · 492K / 1atm = 14.130L 3. The change of internal energy of ideal gas is given by: ?U = n·Cv·?T For an ideal gas Cp - Cv = R with ? = Cp/Cv Cv = R/(?-1) ?U12= n·[R/(?-1)]·(T2 - T1) = 0.35mol · [8.314472J/molK/(1.4-1) ] · (600K - 300K) = 2182.5J The work done one the gas is: W12 = - ?V1?V2 p dV = 0 because the volume does not change Therefore ?U12 = Q12 + W12 => Q12 = ?U12 - W12 = 2182.5J 4. The work done one the gas is: W23 = - ?V2?V3 p dV for an adiabatic process p·V^? = const p·V^? = p2·V2^? p = p2·V2^? ·V^-? Hence: W23 = - ?V2?V3 p2·V2^? ·V^-? dV = - p2·V2^? · ?V2?V3 V^-? dV = - p2·V2^? · (1/(1-?)) · [ V3^(1-?) - V2^(1-?)] = - p2·V2 · (1/(?-1)) · [1 - (V2/V3)^(?-1)] = - p2·V2 · (1/(?-1)) · [1 - (T2/T3)^(?-1)] here calculate in SI units because Pam³=J = - 2·101325Pa · 0.008616m³ · (1/(1.4-1)) · [1 - (300K / 492K )^(1.4-1)] = -783.7J for adiabatic process: Q23 = 0 Hence ?U23 = W23 = -783.7J 5. ?U31= n·[R/(?-1)]·(T1 - T3) = 0.35mol · [8.314472J/molK/(1.4-1) ] · (300K - 492K) = -3990.9J W31 = - ?V3?V1 p dV = 0 because p=constant = p3=p1 W31 = - p1·(V1 - V3) = p1·(V3 - V1) = 101325Pa · (0.014130m³ - 0.008616m³) = 558.7J Q31 = ?U31 - W31 = -3990.9J - 558.7J = -4549.6J