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During isothermal expansion of an ideal gas, its
- a)internal energy increases [1991, 94]
- b)enthalpy decreases
- c)enthalpy remains unaffected
- d)enthalpy reduces to zero.
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
During isothermal expansion of an ideal gas, itsa)internal energy incr...
During isothermal expansion of ideal gas,
ΔT = 0. Now H = E + PV
∵ ΔH = ΔE + Δ(PV)
∴ ΔH = ΔE + Δ(nRTT));
Thus if ΔT = 0., ΔH = ΔE i.e., remain unaffected
ΔT = 0. Now H = E + PV
∵ ΔH = ΔE + Δ(PV)
∴ ΔH = ΔE + Δ(nRTT));
Thus if ΔT = 0., ΔH = ΔE i.e., remain unaffected
Most Upvoted Answer
During isothermal expansion of an ideal gas, itsa)internal energy incr...
Introduction:
In this question, we are asked to determine the effect of isothermal expansion of an ideal gas on its internal energy and enthalpy. Isothermal expansion refers to a process in which the temperature of the gas remains constant throughout the expansion.
Explanation:
Internal Energy:
The internal energy of an ideal gas depends only on its temperature and is independent of its volume. Therefore, during an isothermal expansion where the temperature remains constant, the internal energy of the gas does not change.
Enthalpy:
Enthalpy is defined as the sum of the internal energy and the product of pressure and volume of the gas. Mathematically, enthalpy (H) = internal energy (U) + pressure (P) * volume (V).
During an isothermal expansion, the volume of the gas increases while the temperature remains constant. As a result, the product of pressure and volume (PV) decreases. However, since the internal energy (U) remains constant, the decrease in PV is compensated by an increase in U, keeping the enthalpy constant.
Therefore, during an isothermal expansion of an ideal gas, the enthalpy remains unaffected. This is represented by option 'C' as the correct answer.
Summary:
During an isothermal expansion of an ideal gas, the internal energy remains constant while the enthalpy remains unaffected. The increase in internal energy compensates for the decrease in the product of pressure and volume, resulting in no change in enthalpy.
In this question, we are asked to determine the effect of isothermal expansion of an ideal gas on its internal energy and enthalpy. Isothermal expansion refers to a process in which the temperature of the gas remains constant throughout the expansion.
Explanation:
Internal Energy:
The internal energy of an ideal gas depends only on its temperature and is independent of its volume. Therefore, during an isothermal expansion where the temperature remains constant, the internal energy of the gas does not change.
Enthalpy:
Enthalpy is defined as the sum of the internal energy and the product of pressure and volume of the gas. Mathematically, enthalpy (H) = internal energy (U) + pressure (P) * volume (V).
During an isothermal expansion, the volume of the gas increases while the temperature remains constant. As a result, the product of pressure and volume (PV) decreases. However, since the internal energy (U) remains constant, the decrease in PV is compensated by an increase in U, keeping the enthalpy constant.
Therefore, during an isothermal expansion of an ideal gas, the enthalpy remains unaffected. This is represented by option 'C' as the correct answer.
Summary:
During an isothermal expansion of an ideal gas, the internal energy remains constant while the enthalpy remains unaffected. The increase in internal energy compensates for the decrease in the product of pressure and volume, resulting in no change in enthalpy.
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Community Answer
During isothermal expansion of an ideal gas, itsa)internal energy incr...
If ∆H=∆E+nRdT
where, dT=0
So, ∆H=∆E
where, dT=0
So, ∆H=∆E
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During isothermal expansion of an ideal gas, itsa)internal energy increases [1991, 94]b)enthalpy decreasesc)enthalpy remains unaffectedd)enthalpy reduces to zero.Correct answer is option 'C'. Can you explain this answer?
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