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One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. If the work done during the process is 3 kJ, then final temperature of the gas is (Cv = 20 J/K)
- a)100 K
- b)150 K
- c)195 K
- d)255 K
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
One mole of an ideal gas is allowed to expand reversibly and adiabati...
Since the gas expands adiabatically (i.e., no change in enthalpy) so the heat is totally converted into work. For the gas, Cv = 20 J/K. Thus, 20 J of heat is required for 1° change in temperature of the gas.
Heat change involved during the process i.e., work done = 3kJ = 3000J.
Change in temperature = 3000/20 K = 150 K
Initial temperature = 300 K
Since, the gas expands so the temperature decreases and thus final temperature is
300 - 150 = 150K
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One mole of an ideal gas is allowed to expand reversibly and adiabati...
Given data:
- Initial temperature (T1) = 27°C = 300 K
- Work done (W) = 3 kJ = 3000 J
- Cv = 20 J/K
Concept:
The process is reversible and adiabatic, which means there is no heat exchange with the surroundings (Q = 0) and the entropy remains constant (ΔS = 0). From the first law of thermodynamics, we know that ΔU = Q - W, where ΔU is the change in internal energy. Since Q = 0, ΔU = -W.
Calculation:
1. We know that ΔU = CvΔT, where ΔT is the change in temperature. So, we can write -W = CvΔT.
2. Rearranging the equation, we have ΔT = -W/Cv.
3. Substituting the given values, we get ΔT = -3000 J / 20 J/K = -150 K.
4. Since the temperature change is negative, it means the final temperature (T2) is lower than the initial temperature.
5. Final temperature (T2) = T1 + ΔT = 300 K - 150 K = 150 K.
Therefore, the final temperature of the gas is 150 K, which corresponds to option 'B'.
- Initial temperature (T1) = 27°C = 300 K
- Work done (W) = 3 kJ = 3000 J
- Cv = 20 J/K
Concept:
The process is reversible and adiabatic, which means there is no heat exchange with the surroundings (Q = 0) and the entropy remains constant (ΔS = 0). From the first law of thermodynamics, we know that ΔU = Q - W, where ΔU is the change in internal energy. Since Q = 0, ΔU = -W.
Calculation:
1. We know that ΔU = CvΔT, where ΔT is the change in temperature. So, we can write -W = CvΔT.
2. Rearranging the equation, we have ΔT = -W/Cv.
3. Substituting the given values, we get ΔT = -3000 J / 20 J/K = -150 K.
4. Since the temperature change is negative, it means the final temperature (T2) is lower than the initial temperature.
5. Final temperature (T2) = T1 + ΔT = 300 K - 150 K = 150 K.
Therefore, the final temperature of the gas is 150 K, which corresponds to option 'B'.
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One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. If the work done during the process is 3 kJ, then final temperature of the gas is (Cv = 20 J/K)a)100 Kb)150 Kc)195 Kd)255 KCorrect answer is option 'B'. Can you explain this answer?
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