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Find the final temperature of one mole of an ideal gas at an initial temperature to t K.The gas does 9 R joules of work adiabatically. The ratio of specific heats of this gas at constant pressure and at constant volume is 4/3.
  • a)
    (t-9)K
  • b)
    (t - 4/3)K
  • c)
    t + 3K
  • d)
    (t - 3)K
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Find the final temperature of one mole of an ideal gas at an initial t...
TInitial  = t K
Work, W = 9R
Ratio of specific heats, γ = C/ Cv = 4/3
In an adiabatic process, we have
W = R(TFinal – Tinitial) / (1-γ)
9R = R (TFinal – t) / (1 – 4/3)
TFinal – t = 9 (-1/3) = -3
TFinal  = (t-3) K
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Most Upvoted Answer
Find the final temperature of one mole of an ideal gas at an initial t...
Adiabatic Process:
An adiabatic process is a thermodynamic process in which there is no transfer of heat or mass (matter) between a thermodynamic system and its surroundings.

Work done by the gas:
The gas does 9 R joules of work adiabatically.

Ratio of specific heats:
The ratio of specific heats of this gas at constant pressure and at constant volume is 4/3.

Formula for adiabatic work:
W = (γ - 1) * n * R * (T₁ - T₂) / γ

where,
W = work done
n = number of moles of gas
R = gas constant
T₁ = initial temperature
T₂ = final temperature
γ = ratio of specific heats

Substituting the given values in the above formula, we get

9R = (4/3 - 1) * 1 * R * (t - T₂) / (4/3)

Simplifying, we get

9 = (1/3) * (t - T₂) / (4/3)

27/4 = t - T₂

T₂ = t - 27/4

Final temperature of the gas:
T₂ = t - 27/4
T₂ = t - 6.75
T₂ = (t - 3) K

Therefore, the correct answer is option 'D' which is the final temperature of the gas is (t - 3) K.
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Community Answer
Find the final temperature of one mole of an ideal gas at an initial t...
Work, W = 9R
Ratio of specific heats, γ = Cp / Cv = 4/3
In an adiabatic process, we have
W = R(TFinal – Tinitial) / (1-γ)
9R = R (TFinal – t) / (1 – 4/3)
TFinal – t = 9 (-1/3) = -3
TFinal  = (t-3) K
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Find the final temperature of one mole of an ideal gas at an initial temperature to t K.The gas does 9 R joules of work adiabatically. The ratio of specific heats of this gas at constant pressure and at constant volume is 4/3.a)(t-9)Kb)(t - 4/3)Kc)t + 3Kd)(t - 3)KCorrect answer is option 'D'. Can you explain this answer?
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