3 liters of N2 at 0 oC and 10 atm pressure are expanded Isothermally against a constant pressure 2 atm until the pressure of gas is 5 atm. Assuming the gas to be ideal, what are the value of w, AE, AH dan q for the system
1. 3 liters of N2 at 0 oC and 10 atm pressure are expanded Isothermally against a constant pressure 2 atm until the pressure of gas is 5 atm. Assuming the gas to be ideal, what are the value of w, AE, AH dan q for the system
Jawaban:
w = 4.615 litersatm
AE = 0
AH = 4.615 litersatm
q = 9.230 liters*atm
Penjelasan:
The expansion of the gas is isothermal, meaning that the temperature of the gas remains constant during the expansion. This means that the change in internal energy, AE, of the gas is zero.
The work done by the gas, w, is the product of the pressure of the gas and the change in volume of the gas. The initial volume of the gas can be calculated from the ideal gas law:
PV = nRT
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas.
Substituting the known values, we get:
(10 atm)(V) = (3 liters)(0.08206 Latm/molK)(273.15 K)
Solving for V, we find that the initial volume of the gas is 0.923 liters.
The final volume of the gas can be calculated in the same way, using the final pressure of 5 atm:
(5 atm)(V) = (3 liters)(0.08206 Latm/molK)(273.15 K)
Solving for V, we find that the final volume of the gas is 1.846 liters.
The work done by the gas is then:
w = (5 atm)(1.846 liters - 0.923 liters) = 4.615 liters*atm
The change in enthalpy, AH, of the gas is equal to the change in internal energy plus the work done by the gas. Since the change in internal energy is zero, the change in enthalpy is also equal to the work done by the gas:
AH = w = 4.615 liters*atm
The heat transfer, q, for the system is the change in enthalpy of the gas plus the work done by the gas:
q = AH + w = 4.615 litersatm + 4.615 litersatm = 9.230 liters*atm
Thus, the values of w, AE, AH, and q for the system are:
w = 4.615 litersatm
AE = 0
AH = 4.615 litersatm
q = 9.230 liters*atm
2. Two moles of aluminum consist of...... atoms (L = 6.02 x 10^23)
x = n x L
x = 2 x 6,02 x 10^23
x = 12,02 x 10^23
x = 1,202 x 10^24
3. Two moles of aluminum consist of...... atoms (L = 6.02 x 10^23)
x = n x L
x = 2 x 6,02 x 10^23
x = 12,02 x 10^23
x = 1,202 x 10^24
4. How many moles are there in44g of CO2? How many molecus is this
simply the molar mass of co2 is 44g/mol. It is calculated as below;
molar mass of c = 12 g/mol
molar mass of o= 16 g/mol
molar mass of co2= molar mass of c+ 2(molar mass of o)= (12+16+16)g/mol=44g/mol
number of moles = the given mass / molar mass of co2
number of moles= 44 g / 44 g/mol =1 mol
therfore 44g of co2 have 1 mole of co2 in it.
5. -Q*Today is wednesday, what day was yesterday?*Today is wednesday, tomorrow will be...nt: cool because of moles '_'
Jawaban:
Tuesday
Thursday
thank u '-'
Answer:
1). Today is wednesday, what day was yesterday?
answer: Tuesday
2). Today is wednesday, tomorrow will be...
answer: Thursday
explanation:
hm _-
hope it's useful^^
6. a mixture of gases is analyzed and found to have the following composition:CO2 = 12.0%CH4 = 27.3%H2 = 9.9%N2 = 44.8%CO = 6.0%How much will the weight of 3lb moles of this gas mixture?
Jawaban:
ambil point aja kak
7. Eight moles of helium gas with an initial volume of 50 liters and a pressure of 6 atmospheres are isobarically compressed to a final gas temperature of -150 degrees Celsius. Determine the volume of gas and heat that must be eliminated from the system if the Cpm of gas is 22 J/mol.K.
Jawaban:
The heat that must be eliminated from the system is 1521.2 J.
Penjelasan:
To solve this problem, we need to use the ideal gas law, which states that the volume of a gas is directly proportional to its temperature (measured in Kelvins) and inversely proportional to its pressure. The ideal gas law is given by:
PV = nRT
where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature of the gas in Kelvins.
In this case, we are given that the pressure of the gas is 6 atmospheres, the number of moles of the gas is 8, and the gas constant is 8.314 J/mol.K. The initial temperature of the gas is not given, but we are told that the gas is being compressed to a final temperature of -150 degrees Celsius. To use the ideal gas law, we need to convert this temperature to Kelvins by adding 273.15 to the temperature in Celsius:
Tfinal (in Kelvins) = -150 + 273.15 = 123.15 K
We can now use the ideal gas law to calculate the final volume of the gas:
Vfinal = (nRT) / P = (8 * 8.314 * 123.15) / 6 = 205.7 liters
To determine the heat that must be eliminated from the system, we can use the equation for isobaric heat capacity, which is given by:
q = Cpm * (Tfinal - Tinitial)
where q is the heat that must be eliminated from the system, Cpm is the heat capacity of the gas at constant pressure, Tfinal is the final temperature of the gas, and Tinitial is the initial temperature of the gas. In this case, we are given that Cpm is 22 J/mol.K. We need to determine the initial temperature of the gas in order to calculate the heat that must be eliminated from the system.
To do this, we can use the ideal gas law to solve for the initial volume of the gas, given that the initial pressure and number of moles of the gas are known:
Vinitial = (nRT) / P = (8 * 8.314 * Tinitial) / 6
Substituting the known values, we get:
Vinitial = (8 * 8.314 * Tinitial) / 6 = 50 liters
Solving for Tinitial, we find that the initial temperature of the gas is approximately -217.6 degrees Celsius. Converting this temperature to Kelvins, we find that the initial temperature of the gas is 55.55 K.
Now that we know the initial and final temperatures of the gas, we can use the equation for isobaric heat capacity to calculate the heat that must be eliminated from the system:
q = Cpm * (Tfinal - Tinitial) = 22 * (123.15 - 55.55) = 22 * 67.6 = 1521.2 J
Therefore, the volume of the gas after compression is 205.7 liters, and the heat that must be eliminated from the system is 1521.2 J.
8. An ideal gas 1mol is kept at 0C during expansion from 2L to 8L . Find the amount of work done on the gas during expansion
Jawaban:
Gas ideal 1 mol dijaga pada 0C selama ekspansi dari 2L ke 8L. Hitunglah usaha yang dilakukan gas selama pemuaian.
9. what is the main ideal of paragrap 3
Jawaban:
have lost of hobbies.I like music mostly classical and folk music
semoga membantu
what is the main ideal of paragrap 3
Hobbies
Jangan lupa jadikan jawaban terbaik ✔️
10. The number of moles of solute present in 1 kg of a solvent is called its A. Molality B. Molarity C. Normality D. Formality
A. Molality
pembahasan :
pertanyaannya adalah jumlah mol di larutan yang ada pada 1 kg pelarut disebut apa, jawabannya ada molalitas yaitu (n). sedangkan, molaritas (M) adalah konsentrasi suatu zat pada larutan tertentu.
11. Consider the reaction for the commercial preparation of ammonia ; synthesis of ammonia (NH3) from the reaction between nitrogen and hydrogen. a) How many moles of hydrogen are needed to prepare 312 moles of ammonia? b) How many grams of hydrogen is this?
11010292928822
Penjelasan:
soriy ttayqyquq
12. How many moles of sodium bromide can be formed from 1.55 moles of aluminum bromide?
Jawaban:
12020729629101
Penjelasan:
kalo salah maap
13. Two moles of aluminum consist of...... atoms (L = 6.02 x 10^23)
x = n x L
x = 2 x 6,02 x 10^23
x = 12,02 x 10^23
x = 1,202 x 10^24
14. A mass of an ideal gas of volume v, at pressure p, undergoes a cycle of changes as shown in the graph.
Jawaban:
Sebuah massa gas ideal volume v, pada tekanan p, mengalami siklus perubahan seperti yang ditunjukkan pada grafik.
Penjelasan:
Translate:
A mass of an ideal gas of volume v, at pressure p, undergoes a cycle of changes as shown in the graph.
Sebuah massa gas ideal volume v, pada tekanan p, mengalami siklus perubahan seperti yang ditunjukkan pada grafik.
semoga membantu
15. What is the molarity of a solution containing 4 moles of KCl in 2.5 L of solution?
Penjelasan:
dik : n KCl = 4 mol
v = 2,5 L
dit : M = ?
jb :
M = n / v
= 4 mol / 2,5 L
= 1,6 mol/L
semoga membantu :)
16. For the given reaction C5H12 + 8 O2 → 5 CO2 + 6 H2O, If the reactants C5H12 & O2 are having 2 moles an 8 moles of initial feed respectively. How many moles of C5H12 are there when the reaction completes?
Jawaban:
1 mol C5H12
Penjelasan:
pakai cara yang di gambar ya
semoga bermanfaat
17. If a = -1 and b = 2, the expanded form of 3a + 2b-3 is
Jawab:
3a + 2b-3 = -2
Penjelasan dengan langkah-langkah:
a = -1
b = 2
3a + 2b - 3
= 3(-1) + 2(2) - 3
= (3 x -1) + (2 x 2) - 3
= -3 + 4 - 3
= 1 - 3
= -2
so, the result from 3a + 2b - 3 is -2
18. What is the main ideal of paragraph
it is basically the important point of a paragraf.the topic of each paragraph
19. If a = -1 and b= 3, The expanded form of 3a + 2b – 3 is..
→ 3a + 2b - 3
→ 3(-1) + 2(3) - 3
→ -3 + 6 - 3
→ 3 - 3
→ 0
[tex]\purple{\boxed{\blue{\boxed{\green{\star{\orange{\ \: \: \mathcal{JK} \: \: \: {\green{\star}}}}}}}}} [/tex]
20. how many moles of water is 5.87 x 10 22 molecules ?
Saya jawabnya pakai bahasa indonesia ya :p
Diketahui : Jumlah partikel : 5.87 x [tex]10^{22}[/tex]
Jumlah partikel = mol x L (bilangan avogadro)
5.87 x [tex]10^{22}[/tex] = mol x 6.02 x [tex]10^{23}[/tex]
mol = 0.097 mol
Semoga membantu :D
0 Komentar