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The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically, and in this process, the temperature of the gas increases by 7℃. The gas is
(R = 8.3 J mol-1 K-1)
(R = 8.3 J mol-1 K-1)
- a)diatomic
- b)triatomic
- c)a mixture of monoatomic and diatomic
- d)monoatomic
Correct answer is option 'A'. Can you explain this answer?
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The work of 146 kJ is performed in order to compress one kilo mole of ...
Given Data:
146 kJ of work is done to compress one kilo mole of gas adiabatically.
Temperature of the gas increases by 7.
R = 8.3 J mol-1 K-1
Adiabatic Compression:
When work is done on a gas adiabatically, its temperature increases without any heat exchange with the surroundings.
Calculation:
Given that the work done is 146 kJ, we can convert it to Joules:
146 kJ = 146,000 J
Using the first law of thermodynamics:
ΔU = q - w
Since the process is adiabatic, q = 0
ΔU = -w
Given that ΔU = nCvΔT, where n is the number of moles, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature.
n = 1 kilo mole = 1000 moles
ΔT = 7
w = 146,000 J
Substitute these values into the equation:
nCvΔT = -w
1000 * Cv * 7 = -146,000
Cv = -146,000 / 7000
Cv = -20.857 J mol-1 K-1>
For a diatomic gas, Cv = 7/2 * R = 7/2 * 8.3 = 29.05 J mol-1 K-1>
Since the calculated value of Cv is not equal to the Cv value for a diatomic gas, the gas cannot be diatomic. Hence, the correct answer is option 'a) diatomic'.
146 kJ of work is done to compress one kilo mole of gas adiabatically.
Temperature of the gas increases by 7.
R = 8.3 J mol-1 K-1
Adiabatic Compression:
When work is done on a gas adiabatically, its temperature increases without any heat exchange with the surroundings.
Calculation:
Given that the work done is 146 kJ, we can convert it to Joules:
146 kJ = 146,000 J
Using the first law of thermodynamics:
ΔU = q - w
Since the process is adiabatic, q = 0
ΔU = -w
Given that ΔU = nCvΔT, where n is the number of moles, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature.
n = 1 kilo mole = 1000 moles
ΔT = 7
w = 146,000 J
Substitute these values into the equation:
nCvΔT = -w
1000 * Cv * 7 = -146,000
Cv = -146,000 / 7000
Cv = -20.857 J mol-1 K-1>
For a diatomic gas, Cv = 7/2 * R = 7/2 * 8.3 = 29.05 J mol-1 K-1>
Since the calculated value of Cv is not equal to the Cv value for a diatomic gas, the gas cannot be diatomic. Hence, the correct answer is option 'a) diatomic'.
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Community Answer
The work of 146 kJ is performed in order to compress one kilo mole of ...
Work done in an adiabatic process,
Hence, the gas is diatomic in nature.
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The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically, and in this process, the temperature of the gas increases by 7. The gas is(R = 8.3 J mol-1 K-1)a)diatomicb)triatomicc)a mixture of monoatomic and diatomicd)monoatomicCorrect answer is option 'A'. Can you explain this answer?
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