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In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 is
Correct answer is '9'. Can you explain this answer?
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In a constant volume calorimeter, 3.5 g of a gas with molecular weigh...
Given data:
- Mass of gas = 3.5 g
- Molecular weight of gas = 28
- Initial temperature of calorimeter = 298.0 K
- Final temperature of calorimeter = 298.45 K
- Heat capacity of calorimeter = 2.5 kJ K⁻¹
Calculating the heat absorbed by the calorimeter:
The heat absorbed by the calorimeter can be calculated using the formula:
Q = mcΔT
Where:
Q = heat absorbed by the calorimeter
m = mass of the gas
c = heat capacity of the calorimeter
ΔT = change in temperature of the calorimeter
Plugging in the values, we get:
Q = (3.5 g) * (2.5 kJ K⁻¹) * (298.45 K - 298.0 K)
Q = 0.4375 kJ
Calculating the moles of gas burnt:
The number of moles of gas burnt can be calculated using the formula:
n = mass / molecular weight
Plugging in the values, we get:
n = 3.5 g / 28 g mol⁻¹
n = 0.125 mol
Calculating the enthalpy of combustion:
The enthalpy of combustion can be calculated using the formula:
ΔH = Q / n
Plugging in the values, we get:
ΔH = 0.4375 kJ / 0.125 mol
ΔH = 3.5 kJ mol⁻¹
However, the answer provided is '9' kJ mol⁻¹. It seems that there might be an error in the calculation or the given answer. Please double-check the calculations or clarify if there are any additional considerations to be made.
- Mass of gas = 3.5 g
- Molecular weight of gas = 28
- Initial temperature of calorimeter = 298.0 K
- Final temperature of calorimeter = 298.45 K
- Heat capacity of calorimeter = 2.5 kJ K⁻¹
Calculating the heat absorbed by the calorimeter:
The heat absorbed by the calorimeter can be calculated using the formula:
Q = mcΔT
Where:
Q = heat absorbed by the calorimeter
m = mass of the gas
c = heat capacity of the calorimeter
ΔT = change in temperature of the calorimeter
Plugging in the values, we get:
Q = (3.5 g) * (2.5 kJ K⁻¹) * (298.45 K - 298.0 K)
Q = 0.4375 kJ
Calculating the moles of gas burnt:
The number of moles of gas burnt can be calculated using the formula:
n = mass / molecular weight
Plugging in the values, we get:
n = 3.5 g / 28 g mol⁻¹
n = 0.125 mol
Calculating the enthalpy of combustion:
The enthalpy of combustion can be calculated using the formula:
ΔH = Q / n
Plugging in the values, we get:
ΔH = 0.4375 kJ / 0.125 mol
ΔH = 3.5 kJ mol⁻¹
However, the answer provided is '9' kJ mol⁻¹. It seems that there might be an error in the calculation or the given answer. Please double-check the calculations or clarify if there are any additional considerations to be made.
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Community Answer
In a constant volume calorimeter, 3.5 g of a gas with molecular weigh...
Given, CV = 2.5 kJ K-1 = 2500 J K-1
ΔT = T2 - T1 = 298.45 - 298 = 0.45 K
ΔH due to combustion of 3.5 g gas = CV x ΔT = 2500 x 0.45 = 1125 J
Given, molecular weight of gas = 28.
28 1 mole
Hence, ΔH due to combustion of 1 mole of gas
= 28 = 9000 J
∴ Δ H in kJ mol-1 = 9kJ mol-1.
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In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer?
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In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer?.
In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer?.
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In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer?, a detailed solution for In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer? has been provided alongside types of In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 isCorrect answer is '9'. Can you explain this answer? theory, EduRev gives you an
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