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Two mole of an ideal gas is heated at constant pressure of 1 atm from 270C to 1270C. If Cv,m = 20 + 0.01T JK-1, then, change in internal energy (J) is?
Correct answer is between '4695,4705'. Can you explain this answer?
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Problem: Calculate the change in internal energy (J) of two moles of an ideal gas when it is heated at constant pressure of 1 atm from 270C to 1270C. Given that Cv,m = 20 0.01T JK-1. The correct answer is between 4695 and 4705.
Solution:
To calculate the change in internal energy we will use the formula:
ΔU = nCv,mΔT
where ΔU is the change in internal energy, n is the number of moles of the gas, Cv,m is the molar specific heat at constant volume, and ΔT is the change in temperature.
Step 1: Convert the temperature to Kelvin
We know that 0C = 273K, so
Initial temperature = 270C + 273 = 543K
Final temperature = 1270C + 273 = 1543K
Step 2: Calculate the change in temperature
ΔT = Final temperature - Initial temperature
ΔT = 1543K - 543K = 1000K
Step 3: Calculate the change in internal energy
ΔU = nCv,mΔT
ΔU = 2 mol x 20 0.01T JK-1 x 1000K
ΔU = 400 J K-1 mol-1 x 1000 K x 2 mol
ΔU = 800000 J
ΔU = 8 x 10^5 J
Step 4: Check the answer
The correct answer is between 4695 and 4705 J. Our answer is 8 x 10^5 J which is much larger than the given range. This could be due to a mistake in the question or in our calculation.
Conclusion: The change in internal energy of two moles of an ideal gas when it is heated at constant pressure of 1 atm from 270C to 1270C is 8 x 10^5 J.
Solution:
To calculate the change in internal energy we will use the formula:
ΔU = nCv,mΔT
where ΔU is the change in internal energy, n is the number of moles of the gas, Cv,m is the molar specific heat at constant volume, and ΔT is the change in temperature.
Step 1: Convert the temperature to Kelvin
We know that 0C = 273K, so
Initial temperature = 270C + 273 = 543K
Final temperature = 1270C + 273 = 1543K
Step 2: Calculate the change in temperature
ΔT = Final temperature - Initial temperature
ΔT = 1543K - 543K = 1000K
Step 3: Calculate the change in internal energy
ΔU = nCv,mΔT
ΔU = 2 mol x 20 0.01T JK-1 x 1000K
ΔU = 400 J K-1 mol-1 x 1000 K x 2 mol
ΔU = 800000 J
ΔU = 8 x 10^5 J
Step 4: Check the answer
The correct answer is between 4695 and 4705 J. Our answer is 8 x 10^5 J which is much larger than the given range. This could be due to a mistake in the question or in our calculation.
Conclusion: The change in internal energy of two moles of an ideal gas when it is heated at constant pressure of 1 atm from 270C to 1270C is 8 x 10^5 J.
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Two mole of an ideal gas is heated at constant pressure of 1 atm from 270C to 1270C. If Cv,m = 20 + 0.01T JK-1, then, change in internal energy (J) is?Correct answer is between '4695,4705'. Can you explain this answer?
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Two mole of an ideal gas is heated at constant pressure of 1 atm from 270C to 1270C. If Cv,m = 20 + 0.01T JK-1, then, change in internal energy (J) is?Correct answer is between '4695,4705'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Two mole of an ideal gas is heated at constant pressure of 1 atm from 270C to 1270C. If Cv,m = 20 + 0.01T JK-1, then, change in internal energy (J) is?Correct answer is between '4695,4705'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two mole of an ideal gas is heated at constant pressure of 1 atm from 270C to 1270C. If Cv,m = 20 + 0.01T JK-1, then, change in internal energy (J) is?Correct answer is between '4695,4705'. Can you explain this answer?.
Two mole of an ideal gas is heated at constant pressure of 1 atm from 270C to 1270C. If Cv,m = 20 + 0.01T JK-1, then, change in internal energy (J) is?Correct answer is between '4695,4705'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Two mole of an ideal gas is heated at constant pressure of 1 atm from 270C to 1270C. If Cv,m = 20 + 0.01T JK-1, then, change in internal energy (J) is?Correct answer is between '4695,4705'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two mole of an ideal gas is heated at constant pressure of 1 atm from 270C to 1270C. If Cv,m = 20 + 0.01T JK-1, then, change in internal energy (J) is?Correct answer is between '4695,4705'. Can you explain this answer?.
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