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
An isotope is when an element has the same number of protons and different number of neutrons. That is why the atomic mass changes.
Answer:
N2 is lost from the craft 6.9% faster than O2 is lost.
Explanation:
<u>Step 1:</u> Data given
0.500 atm of N2
0.500 atm of 02
Molar mass of 02 = 2*16 g/mole = 32 g/mole
Molar mass of N2 = 2* 14 g/mole = 28g/mole
<u>Step 2:</u> Calculate the rate of loss
r1N2 / r2O2 = √(M2(02)/M1(N2)) = √(32/28) = 1.069
1.069-1)*100= 6.9%
This means N2 is lost from the craft 6.9% faster than O2 is lost.
"<span>A. The polar regions have long winters and short, cool summers." would be the correct answer.
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Answer:
pH = 9.6
Explanation:
According to Brönsted-Lowry theory, NH₃ is a base and NH₄⁺ its conjugate acid. When they are together in a solution, the form a buffer, which is used to resist abrupt changes in pH when an acid or a base is added. pOH fro a buffer can be found using Henderson-Hasselbalch equation.
Since NH₄Cl is a strong electrolyte, [NH₄Cl] = [NH₄⁺]
![pOH = pKb + log\frac{[NH_{4}^{+} ]}{[NH_{3}]} =4.7+log\frac{0.035M}{0.070M} =4.4](https://tex.z-dn.net/?f=pOH%20%3D%20pKb%20%2B%20log%5Cfrac%7B%5BNH_%7B4%7D%5E%7B%2B%7D%20%5D%7D%7B%5BNH_%7B3%7D%5D%7D%20%3D4.7%2Blog%5Cfrac%7B0.035M%7D%7B0.070M%7D%20%3D4.4)
Now, we can find pH using the following expression:
pH + pOH = 14
pH = 14 - pOH = 14 - 4.4 = 9.6
