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1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to (Given: CV = 20 J mol–1K–1):
Correct answer is '150'. Can you explain this answer?
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1 mole of an ideal gas is allowed to expand reversibly and adiabatical...
For adiabatic process, q=0. So, ∆U=W.Now, Cv=dU/dT. So, dU=Cv*dT and hence W=Cv*dT. (2)Now, W=3kJ=3000J.So, Change in temperature=dT= W/Cv = 150 K.Initial temperature=27DEGC=300 K.So, Final temperature= 300-150 (since,dT=150) = 150
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1 mole of an ideal gas is allowed to expand reversibly and adiabatical...
The temperature of the gas is given as 27°C, which can be converted to Kelvin by adding 273.15:
T = 27 + 273.15 = 300.15 K
Since the process is adiabatic, it means that there is no heat exchange with the surroundings, so q = 0.
The change in entropy (ΔS) for an adiabatic process is given by the equation:
ΔS = C_v * ln(T2/T1)
Where:
ΔS = change in entropy
C_v = molar heat capacity at constant volume
T1 = initial temperature
T2 = final temperature
For an ideal gas, the molar heat capacity at constant volume is given by the equation:
C_v = (3/2)R
Where:
R = ideal gas constant = 8.314 J/(mol·K)
Substituting the values into the equation:
C_v = (3/2) * 8.314 J/(mol·K) = 12.471 J/(mol·K)
ΔS = 12.471 * ln(T2/T1)
To find the final temperature, we need more information about the process (such as the volume or pressure at the end). Without this information, we cannot calculate the final temperature or the change in entropy.
T = 27 + 273.15 = 300.15 K
Since the process is adiabatic, it means that there is no heat exchange with the surroundings, so q = 0.
The change in entropy (ΔS) for an adiabatic process is given by the equation:
ΔS = C_v * ln(T2/T1)
Where:
ΔS = change in entropy
C_v = molar heat capacity at constant volume
T1 = initial temperature
T2 = final temperature
For an ideal gas, the molar heat capacity at constant volume is given by the equation:
C_v = (3/2)R
Where:
R = ideal gas constant = 8.314 J/(mol·K)
Substituting the values into the equation:
C_v = (3/2) * 8.314 J/(mol·K) = 12.471 J/(mol·K)
ΔS = 12.471 * ln(T2/T1)
To find the final temperature, we need more information about the process (such as the volume or pressure at the end). Without this information, we cannot calculate the final temperature or the change in entropy.
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1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to (Given: CV = 20 J mol–1K–1):Correct answer is '150'. Can you explain this answer?
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