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7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is
(Given R = 8.3 JK-1 mol-1)
(Given R = 8.3 JK-1 mol-1)
- a)5,810 J
- b)3,486 J
- c)11,620 J
- d)6,972 J
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
7 moles of certain monoatomic ideal gas undergoes a temperature increa...
For a quasi-static process, the change in internal energy of an ideal gas is:
U = nCV ΔT
[Molar heat capacity at constant volume for monoatomic gas = 3R/2]
Δ
[Molar heat capacity at constant volume for monoatomic gas = 3R/2]
Most Upvoted Answer
7 moles of certain monoatomic ideal gas undergoes a temperature increa...
To find the increase in the internal energy of the gas, we can use the equation:
ΔU = nCvΔT
where ΔU is the change in internal energy, n is the number of moles of gas, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature.
In this case, we are given that there are 7 moles of gas, the temperature increases by 40 K, and we need to find the change in internal energy.
- Calculation:
ΔU = nCvΔT
Substituting the given values:
ΔU = 7 × Cv × 40
To find Cv, we can use the relation:
Cp - Cv = R
where Cp is the molar specific heat at constant pressure and R is the ideal gas constant.
Rearranging the equation, we get:
Cv = Cp - R
Given that R = 8.3 J/K/mol, we need to find Cp to calculate Cv.
Using the relation:
Cp = Cv + R
Substituting the value of R, we get:
Cp = Cv + 8.3
Now, substituting the value of Cp in the equation for ΔU, we have:
ΔU = 7 × (Cv + 8.3) × 40
Simplifying the equation:
ΔU = 7 × (Cv × 40 + 8.3 × 40)
ΔU = 7 × (Cv × 40 + 332)
Now, we can substitute the value of Cv = Cp - R:
ΔU = 7 × ((Cp - R) × 40 + 332)
ΔU = 7 × (Cp × 40 - 40R + 332)
Since we know that Cp - Cv = R, we can substitute Cp - R for Cv in the equation:
ΔU = 7 × (Cp × 40 - 40R + 332)
ΔU = 7 × (Cp × 40 - 40(Cp - R) + 332)
ΔU = 7 × (Cp × 40 - 40Cp + 40R + 332)
ΔU = 7 × (-39Cp + 40R + 332)
Finally, substituting the value of R = 8.3 J/K/mol, we get:
ΔU = 7 × (-39Cp + 40(8.3) + 332)
ΔU = 7 × (-39Cp + 332 + 332)
ΔU = 7 × (-39Cp + 664)
Now we can see that the change in internal energy is proportional to -39Cp + 664. Since Cp is positive, the change in internal energy will be negative.
Comparing the given options, we find that option B (-3,486 J) is the closest value to -39Cp + 664. Therefore, the correct answer is option B.
ΔU = nCvΔT
where ΔU is the change in internal energy, n is the number of moles of gas, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature.
In this case, we are given that there are 7 moles of gas, the temperature increases by 40 K, and we need to find the change in internal energy.
- Calculation:
ΔU = nCvΔT
Substituting the given values:
ΔU = 7 × Cv × 40
To find Cv, we can use the relation:
Cp - Cv = R
where Cp is the molar specific heat at constant pressure and R is the ideal gas constant.
Rearranging the equation, we get:
Cv = Cp - R
Given that R = 8.3 J/K/mol, we need to find Cp to calculate Cv.
Using the relation:
Cp = Cv + R
Substituting the value of R, we get:
Cp = Cv + 8.3
Now, substituting the value of Cp in the equation for ΔU, we have:
ΔU = 7 × (Cv + 8.3) × 40
Simplifying the equation:
ΔU = 7 × (Cv × 40 + 8.3 × 40)
ΔU = 7 × (Cv × 40 + 332)
Now, we can substitute the value of Cv = Cp - R:
ΔU = 7 × ((Cp - R) × 40 + 332)
ΔU = 7 × (Cp × 40 - 40R + 332)
Since we know that Cp - Cv = R, we can substitute Cp - R for Cv in the equation:
ΔU = 7 × (Cp × 40 - 40R + 332)
ΔU = 7 × (Cp × 40 - 40(Cp - R) + 332)
ΔU = 7 × (Cp × 40 - 40Cp + 40R + 332)
ΔU = 7 × (-39Cp + 40R + 332)
Finally, substituting the value of R = 8.3 J/K/mol, we get:
ΔU = 7 × (-39Cp + 40(8.3) + 332)
ΔU = 7 × (-39Cp + 332 + 332)
ΔU = 7 × (-39Cp + 664)
Now we can see that the change in internal energy is proportional to -39Cp + 664. Since Cp is positive, the change in internal energy will be negative.
Comparing the given options, we find that option B (-3,486 J) is the closest value to -39Cp + 664. Therefore, the correct answer is option B.
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7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is(Given R = 8.3 JK-1mol-1)a)5,810 Jb)3,486 Jc)11,620 Jd)6,972 JCorrect answer is option 'B'. Can you explain this answer?
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7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is(Given R = 8.3 JK-1mol-1)a)5,810 Jb)3,486 Jc)11,620 Jd)6,972 JCorrect answer is option 'B'. 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 7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is(Given R = 8.3 JK-1mol-1)a)5,810 Jb)3,486 Jc)11,620 Jd)6,972 JCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is(Given R = 8.3 JK-1mol-1)a)5,810 Jb)3,486 Jc)11,620 Jd)6,972 JCorrect answer is option 'B'. Can you explain this answer?.
7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is(Given R = 8.3 JK-1mol-1)a)5,810 Jb)3,486 Jc)11,620 Jd)6,972 JCorrect answer is option 'B'. 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 7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is(Given R = 8.3 JK-1mol-1)a)5,810 Jb)3,486 Jc)11,620 Jd)6,972 JCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is(Given R = 8.3 JK-1mol-1)a)5,810 Jb)3,486 Jc)11,620 Jd)6,972 JCorrect answer is option 'B'. Can you explain this answer?.
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