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2 mol of an ideal gas expanded isothermally and reversibly from 1L to 10L at 300 K. What is the enthalpy change(kJ)?
Correct answer is '0'. Can you explain this answer?
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2 mol of an ideal gas expanded isothermally and reversibly from 1L to ...
Explanation:
Given:
Number of moles of ideal gas, n = 2 mol
Initial volume, V1 = 1 L
Final volume, V2 = 10 L
Temperature, T = 300 K
Isothermal process, ΔT = 0
Reversible process
Formula:
The enthalpy change for an isothermal process is given by:
ΔH = nRTln(V2/V1)
Where,
R = Gas constant = 8.314 J/mol.K
ln = Natural logarithm function
Calculation:
ΔH = nRTln(V2/V1)
= (2 mol)(8.314 J/mol.K)(300 K)ln(10/1)
= 0 J
= 0 kJ
Conclusion:
The enthalpy change for the given isothermal and reversible process is 0 kJ. This is because the temperature remains constant during the process and no energy is exchanged with the surroundings as the process is reversible. Therefore, there is no change in the enthalpy of the system.
Given:
Number of moles of ideal gas, n = 2 mol
Initial volume, V1 = 1 L
Final volume, V2 = 10 L
Temperature, T = 300 K
Isothermal process, ΔT = 0
Reversible process
Formula:
The enthalpy change for an isothermal process is given by:
ΔH = nRTln(V2/V1)
Where,
R = Gas constant = 8.314 J/mol.K
ln = Natural logarithm function
Calculation:
ΔH = nRTln(V2/V1)
= (2 mol)(8.314 J/mol.K)(300 K)ln(10/1)
= 0 J
= 0 kJ
Conclusion:
The enthalpy change for the given isothermal and reversible process is 0 kJ. This is because the temperature remains constant during the process and no energy is exchanged with the surroundings as the process is reversible. Therefore, there is no change in the enthalpy of the system.
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Community Answer
2 mol of an ideal gas expanded isothermally and reversibly from 1L to ...
For isotherml chnge in energy and enthalpy of ideal gas is always zero
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2 mol of an ideal gas expanded isothermally and reversibly from 1L to 10L at 300 K. What is the enthalpy change(kJ)?Correct answer is '0'. Can you explain this answer?
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