One hundred cubic meters of carbon dioxide initially at 150 oC and 50 bar is to be isothermally compressed in a frictionless pis
ton-and-cylinder device to a final pressure of 300 bar. Calculate i. The volume of the compressed gas ii. The work done to compress gas iii. The heat flow on compression assuming carbon dioxide (a) is an ideal gas (b) Obeys principle of corresponding states of section 6.6 (hint: See Illustration 6.6-2) (c) Obeys the Peng-Robinson equation of state (use the visual basic Peng
Answer: hundred cubic meters of carbon dioxide initially at 150◦C and 50 bar is to beisothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar.Calculatei The volume of the compressed gas.ii The work done to compress the gas.iii The heat flow on compression.assuming carbon dioxide(a) Is an ideal gas.(b) Obeys the principle of corresponding states of Sec. 6.6(c) Obeys the Peng-Robinson equation of state.SolutionWe haveT1= 150◦C,P1= 50 bar,T2= 150◦C,P2= 300 bar. (1 and 2 denote the initialand final conditions in this ’snapshot’ problem, respectively - we have sometimes called themt1andt2)(a) If CO2is an idea gas, we havePV=NRT.The number of moles can be calculated from theinitial conditions:N1=P 1 V 1 RT 1 = (6 × 10 6 Pa)(100 m 3 ) (8 . 314 J/(mol K))(150 + 273 K) = 142127 mol = 142 . 1 kmol i. Since we know N 1 = N 2 , T 2 , P 2 V 2 = N 2 RT 2 P 2 = (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) 30 × 10 6 Pa = 16 . 66 m 3 ii. Since there is no shaft work, and since the gas is isothermally compressed , we only have pressure-volume work: W = - Z V 2 V 1 PdV = - Z V 2 V 1 NRT V dV = - NRT ln V 2 V 1 W = - (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) ln 16 . 66 100 = 8 . 958 × 10 8 J iii. Energy balance (integral form) for the closed system is: U 2 - U 1 = Q + W Back from homework 2, for an ideal gas, stating from equation 6.2-21, dU = C V dT + " T ∂P ∂T V - P # dV reduces to: dU = C V dT However, the process is isothermal, so dT = 0 Which gives: dU = Δ U = 1 N Δ U Therefore 0 = Q + W → Q = - W Q = - 8 . 958 × 10 8 J 3
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Explanation:As revealed above, the stimuli connections are clearly stated
Like this? 234+34.1= 268.1 then round. If it is less than 5 then you round down if it is more then you round up. Because it is less the final number would be 268.1=268
A computer's ability to perform so many calculations in a very short time affects how people communicate by stopping them from talking actually. We no longer talk verbally. We don't go out and socialize, we always talk from the phone or text. We have lost our inner touch with talking.
An Olympic decoration is granted to effective contenders at one of the Olympic Amusements. There are three classes of decoration: gold, granted to the victor; silver, granted to the first sprinter up; and bronze, granted to the second sprinter up.
