Below are 3 question I can't seem to get right. I worked them through with my chem tutor and still got them wrong. Any help?
1) A 100 W electric heater (1 W = 1 J/s) operates for 13.5 min to heat the gas in a cylinder.
At the same time, the gas expands from 2 L to 12 L against a constant atmospheric pressure of 3.756 atm. What is the change in internal energy of the gas? Answer in units of kJ
The answer I got was 80.96kJ, that is incorrect.
2) The standard molar internal energy of formation of N2O5(g) is 17.433 kJ/mol at 298 K. What is the standard molar enthalpy of formation of N2O5(g) at the same temperature?
1. 4.955 kJ/mol
2. None of the other answers is correct within 2%.
3. 19.07 kJ/mol
4. 7.921 kJ/mol
5. 11.24 kJ/mol
I chose number 2, which is incorrect.
3) Calculate ∆Suniverse for the condensation of methanol (CH3OH) at 10◦C if the S◦m for CH3OH(g) and CH3OH(ℓ) are 240 J · mol−1·K−1and 127 J·mol−1·K−1, respectively. The enthalpy of vaporization of methanol, ∆Hvap = 37.4 kJ · mol−1.
1. 132 J · mol−1· K−1
2. 113 J · mol−1· K−1
3. −113 J · mol−1· K−1
4. 19 J · mol−1· K−1
5. −132 J · mol−1· K−1
6. −19 J · mol−1· K−1
Both 2 and 3 are incorrect.
I would appreciate any help, my exam is Monday. Thank you
1) A 100 W electric heater (1 W = 1 J/s) operates for 13.5 min to heat the gas in a cylinder.
At the same time, the gas expands from 2 L to 12 L against a constant atmospheric pressure of 3.756 atm. What is the change in internal energy of the gas? Answer in units of kJ
The answer I got was 80.96kJ, that is incorrect.
2) The standard molar internal energy of formation of N2O5(g) is 17.433 kJ/mol at 298 K. What is the standard molar enthalpy of formation of N2O5(g) at the same temperature?
1. 4.955 kJ/mol
2. None of the other answers is correct within 2%.
3. 19.07 kJ/mol
4. 7.921 kJ/mol
5. 11.24 kJ/mol
I chose number 2, which is incorrect.
3) Calculate ∆Suniverse for the condensation of methanol (CH3OH) at 10◦C if the S◦m for CH3OH(g) and CH3OH(ℓ) are 240 J · mol−1·K−1and 127 J·mol−1·K−1, respectively. The enthalpy of vaporization of methanol, ∆Hvap = 37.4 kJ · mol−1.
1. 132 J · mol−1· K−1
2. 113 J · mol−1· K−1
3. −113 J · mol−1· K−1
4. 19 J · mol−1· K−1
5. −132 J · mol−1· K−1
6. −19 J · mol−1· K−1
Both 2 and 3 are incorrect.
I would appreciate any help, my exam is Monday. Thank you
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1) quantity of heat supplied when electric heater operates for 13.5 min or 13.5 x 60 = 810 s ...q = 100 x 810 = 81000 j
W = -Pext(dV) = -Pext (V2-V1) = -3.756 X (12-2) = -3.756 X 10 = -37.56 L atm
1 L atm = 101.3 j
so W = -37.56 X 101.3 j = -3804.828 j
delta U = q + W = 81000 - 3804.828 = 77195.172 j or 77.2 kj
(2) N2 (g) + 5/2O2 (g) -----> N2O5 (g) ....delta H = ?
we know that delta H = delta U + delta ng RT
where delta U = 17.433 kj/mole = 17433 j/mole
delta ng = no.of moles of gaseous products - no. of moles of gaseous reactants = 1 - ( 1 + 5/2) = 1 - 7/2 = -5/2
R = 8.314 J/K/mole
T = 298 K
putting the values...
delta H = 17433 - 5/2 X 8.314 X 298
delta H = 17433 - 6193.93 = 11239.07 j/mole or 11.239 kj/mole or 11.24 kj/mole
so answer is 5.
(3) CH3OH (g) ----> CH3OH (l) ...
delta S (system) = So(CH3OH (l) ) - So (CH3OH (g) ) = 127 - 240 = -113 J /mole/K
now when 1 mole of CH3OH condenses at 10 degree C or 283 k.... 37.4 kj or 37400 j of energy is given to surroundings
so delta S (surr) = 37400 /283 = 132.16 j/mole/K
delta S (universe) = delta S (system) + delta S (surr) = -113 + 132.16 = 19.16 j/mole/K
so answer is 4 .
also please note that delta S (universe) can never be negative due to second law of thermodynamics ....individual delta S (system) or dlta S (surroundigs) can be negative but their sum can never be negative ...
feel free to ask ny question
W = -Pext(dV) = -Pext (V2-V1) = -3.756 X (12-2) = -3.756 X 10 = -37.56 L atm
1 L atm = 101.3 j
so W = -37.56 X 101.3 j = -3804.828 j
delta U = q + W = 81000 - 3804.828 = 77195.172 j or 77.2 kj
(2) N2 (g) + 5/2O2 (g) -----> N2O5 (g) ....delta H = ?
we know that delta H = delta U + delta ng RT
where delta U = 17.433 kj/mole = 17433 j/mole
delta ng = no.of moles of gaseous products - no. of moles of gaseous reactants = 1 - ( 1 + 5/2) = 1 - 7/2 = -5/2
R = 8.314 J/K/mole
T = 298 K
putting the values...
delta H = 17433 - 5/2 X 8.314 X 298
delta H = 17433 - 6193.93 = 11239.07 j/mole or 11.239 kj/mole or 11.24 kj/mole
so answer is 5.
(3) CH3OH (g) ----> CH3OH (l) ...
delta S (system) = So(CH3OH (l) ) - So (CH3OH (g) ) = 127 - 240 = -113 J /mole/K
now when 1 mole of CH3OH condenses at 10 degree C or 283 k.... 37.4 kj or 37400 j of energy is given to surroundings
so delta S (surr) = 37400 /283 = 132.16 j/mole/K
delta S (universe) = delta S (system) + delta S (surr) = -113 + 132.16 = 19.16 j/mole/K
so answer is 4 .
also please note that delta S (universe) can never be negative due to second law of thermodynamics ....individual delta S (system) or dlta S (surroundigs) can be negative but their sum can never be negative ...
feel free to ask ny question