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5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK-1 mol–1, calcuIate Δ U and Δ PV for this process. ( R = 8.0 JK -1 mol-1 )
  • a)
    ΔU = 14 kJ ; Δ( pV ) = 18 kJ
  • b)
    ΔU = 14 J ; Δ( pV ) = 0.8 J
  • c)
    ΔU = 14 kJ ;D( pV ) = 4 kJ
  • d)
    ΔU = 2.8 kJ ; Δ( pV ) = 0.8 kJ
Correct answer is option 'C'. Can you explain this answer?
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5 moles of an ideal gas at 100 K are allowed to undergo reversible com...
ΔU = 28x5x100 = 14kJ 
Δ(pV) = p2v2 - p1v2 = nR (T2 - T1)
= 5x8.314x100 ≈ 4KJ
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5 moles of an ideal gas at 100 K are allowed to undergo reversible com...
To find the work done by the gas during the compression, we can use the equation:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the gas.

Since the process is reversible, we can use the equation:

ΔU = nCvΔT

where n is the number of moles, Cv is the molar heat capacity at constant volume, and ΔT is the change in temperature.

Given that n = 5 moles, Cv = 28 JK^(-1)mol^(-1), ΔT = 200 K - 100 K = 100 K, we can calculate the change in internal energy:

ΔU = 5 mol * 28 JK^(-1)mol^(-1) * 100 K = 14000 J

Since the process is reversible, ΔU = Q - W. Therefore, the work done by the gas during the compression is:

W = Q - ΔU = Q - 14000 J

Since we are not given any information about the heat transfer, we cannot calculate the exact value of the work done by the gas.
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5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK-1mol–1, calcuIate ΔU and ΔPV for this process. ( R = 8.0 JK -1 mol-1 )a)ΔU = 14 kJ ; Δ( pV ) = 18 kJb)ΔU = 14 J ; Δ( pV ) = 0.8 Jc)ΔU = 14 kJ ;D( pV ) = 4 kJd)ΔU = 2.8 kJ ; Δ( pV ) = 0.8 kJCorrect answer is option 'C'. Can you explain this answer?
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5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK-1mol–1, calcuIate ΔU and ΔPV for this process. ( R = 8.0 JK -1 mol-1 )a)ΔU = 14 kJ ; Δ( pV ) = 18 kJb)ΔU = 14 J ; Δ( pV ) = 0.8 Jc)ΔU = 14 kJ ;D( pV ) = 4 kJd)ΔU = 2.8 kJ ; Δ( pV ) = 0.8 kJCorrect answer is option 'C'. 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 5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK-1mol–1, calcuIate ΔU and ΔPV for this process. ( R = 8.0 JK -1 mol-1 )a)ΔU = 14 kJ ; Δ( pV ) = 18 kJb)ΔU = 14 J ; Δ( pV ) = 0.8 Jc)ΔU = 14 kJ ;D( pV ) = 4 kJd)ΔU = 2.8 kJ ; Δ( pV ) = 0.8 kJCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK-1mol–1, calcuIate ΔU and ΔPV for this process. ( R = 8.0 JK -1 mol-1 )a)ΔU = 14 kJ ; Δ( pV ) = 18 kJb)ΔU = 14 J ; Δ( pV ) = 0.8 Jc)ΔU = 14 kJ ;D( pV ) = 4 kJd)ΔU = 2.8 kJ ; Δ( pV ) = 0.8 kJCorrect answer is option 'C'. Can you explain this answer?.
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