1. Given process

An ideal gas undergoes isothermal expansion at constant pressure.

• Isothermal means: $T = \text{constant}$
• Expansion means: volume increases, so entropy is expected to increase.

2. Change in enthalpy for an ideal gas

For an ideal gas, enthalpy depends only on temperature:

$H = H(T)$

Therefore,

$\Delta H = n C_p \Delta T$

Since the process is isothermal,

$\Delta T = 0$

So,

$\Delta H = 0$

Thus, enthalpy remains constant.

3. Change in entropy

For an ideal gas,

$\Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right) - nR \ln\left(\frac{P_2}{P_1}\right)$

Since the process is isothermal,

$T_2 = T_1 \Rightarrow \ln\left(\frac{T_2}{T_1}\right)=0$

Also, pressure is constant, so strictly for an ideal gas, if $T$ and $P$ are both constant, then from

$PV=nRT$

volume cannot change. So the phrase "isothermal expansion at constant pressure" is physically inconsistent for a fixed amount of ideal gas.

However, among the given options, the intended idea is clearly that during expansion the entropy increases, while for an ideal gas at constant temperature enthalpy does not change.

Also, from the common entropy relation for isothermal expansion:

$\Delta S = nR \ln\left(\frac{V_2}{V_1}\right)$

Since expansion implies $V_2 > V_1$,

$\ln\left(\frac{V_2}{V_1}\right) > 0$

Hence,

$\Delta S > 0$

So, entropy increases.

4. Evaluate options

• A: enthalpy increases but entropy decreases → false
• B: enthalpy remains constant but entropy increases → true
• C: enthalpy decreases but entropy increases → false
• D: both enthalpy and entropy remain constant → false

5. Final answer

The correct option is:

$\boxed{\text{B}}$

6. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They match.