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At 270C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of ΔE and q are (R = 2):
- a)0, -965.84 cal
- b)-965.84 cal, -865.58 cal
- c)865.58 cal, -865.58 cal
- d)-865.58 cal, -865.58 cal
Correct answer is 'A'. Can you explain this answer?
Verified Answer
At 270C one mole of an ideal gas is compressed isothermally and revers...
Isothermally (at constant temperature) and reversible work.
at constant temperature , ΔE = 0
ΔE = Q + W,
ΔE = Q + W,
Q = −W = −965.84 cal
Most Upvoted Answer
At 270C one mole of an ideal gas is compressed isothermally and revers...
∆U = 0 because ideal gas haven't potential energy.
q = -w (for isothermal/
work will be positive..
use w = -nRTln(P1/P2)
q = -w (for isothermal/
work will be positive..
use w = -nRTln(P1/P2)
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Community Answer
At 270C one mole of an ideal gas is compressed isothermally and revers...
Understanding the Problem
In this problem, we are dealing with an ideal gas undergoing isothermal and reversible compression. The key parameters are:
- Temperature (T) = 270°C = 543 K (conversion to Kelvin)
- Initial pressure (P1) = 2 atm
- Final pressure (P2) = 10 atm
- Moles of gas (n) = 1
Change in Internal Energy (ΔE)
For an ideal gas undergoing an isothermal process, the change in internal energy (ΔE) is zero. This is because the internal energy of an ideal gas depends only on temperature, and since the temperature remains constant, we can state:
- ΔE = 0
Heat Transfer (q)
In an isothermal process, the heat absorbed or released (q) can be calculated using the formula:
- q = -W (work done on the gas)
To calculate work done (W) in an isothermal process, we use the formula:
- W = nRT ln(P2/P1)
Substituting the values:
- W = 1 mole * 2 (R) * 543 K * ln(10/2)
Calculating gives:
- W ≈ 965.84 cal
Since this work is done on the gas, q will be negative:
- q = -965.84 cal
Final Results
Based on the calculations:
- ΔE = 0
- q = -965.84 cal
Thus, the correct answer is:
- a) ΔE = 0, q = -965.84 cal
In this problem, we are dealing with an ideal gas undergoing isothermal and reversible compression. The key parameters are:
- Temperature (T) = 270°C = 543 K (conversion to Kelvin)
- Initial pressure (P1) = 2 atm
- Final pressure (P2) = 10 atm
- Moles of gas (n) = 1
Change in Internal Energy (ΔE)
For an ideal gas undergoing an isothermal process, the change in internal energy (ΔE) is zero. This is because the internal energy of an ideal gas depends only on temperature, and since the temperature remains constant, we can state:
- ΔE = 0
Heat Transfer (q)
In an isothermal process, the heat absorbed or released (q) can be calculated using the formula:
- q = -W (work done on the gas)
To calculate work done (W) in an isothermal process, we use the formula:
- W = nRT ln(P2/P1)
Substituting the values:
- W = 1 mole * 2 (R) * 543 K * ln(10/2)
Calculating gives:
- W ≈ 965.84 cal
Since this work is done on the gas, q will be negative:
- q = -965.84 cal
Final Results
Based on the calculations:
- ΔE = 0
- q = -965.84 cal
Thus, the correct answer is:
- a) ΔE = 0, q = -965.84 cal
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At 270C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of ΔE and q are (R = 2):a)0, -965.84 calb)-965.84 cal, -865.58 cal c)865.58 cal, -865.58 cald)-865.58 cal, -865.58 calCorrect answer is 'A'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 270C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of ΔE and q are (R = 2):a)0, -965.84 calb)-965.84 cal, -865.58 cal c)865.58 cal, -865.58 cald)-865.58 cal, -865.58 calCorrect answer is 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 270C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of ΔE and q are (R = 2):a)0, -965.84 calb)-965.84 cal, -865.58 cal c)865.58 cal, -865.58 cald)-865.58 cal, -865.58 calCorrect answer is 'A'. Can you explain this answer?.
At 270C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of ΔE and q are (R = 2):a)0, -965.84 calb)-965.84 cal, -865.58 cal c)865.58 cal, -865.58 cald)-865.58 cal, -865.58 calCorrect answer is 'A'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 270C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of ΔE and q are (R = 2):a)0, -965.84 calb)-965.84 cal, -865.58 cal c)865.58 cal, -865.58 cald)-865.58 cal, -865.58 calCorrect answer is 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 270C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of ΔE and q are (R = 2):a)0, -965.84 calb)-965.84 cal, -865.58 cal c)865.58 cal, -865.58 cald)-865.58 cal, -865.58 calCorrect answer is 'A'. Can you explain this answer?.
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