Question

1. calculate the enthalpy change and the heat when 5 moles helium(he) gas underwent a isobaric process from 25°c to 100°c.

Explanation:

Step1: Identify the molar - specific heat at constant pressure for He

For a mono - atomic ideal gas like He, the molar - specific heat at constant pressure $C_p=\frac{5}{2}R$, where $R = 8.314\ J/(mol\cdot K)$.
$C_p=\frac{5}{2}\times8.314\ J/(mol\cdot K)=20.785\ J/(mol\cdot K)$

Step2: Convert temperatures to Kelvin

$T_1 = 25^{\circ}C=(25 + 273.15)K=298.15K$
$T_2 = 100^{\circ}C=(100 + 273.15)K = 373.15K$
$\Delta T=T_2 - T_1=373.15K - 298.15K = 75K$

Step3: Calculate the enthalpy change

The formula for the enthalpy change $\Delta H$ in an isobaric process is $\Delta H=nC_p\Delta T$.
Given $n = 5mol$, $C_p = 20.785\ J/(mol\cdot K)$ and $\Delta T = 75K$
$\Delta H=5mol\times20.785\ J/(mol\cdot K)\times75K=7794.375J$

Step4: Recall the relationship between heat and enthalpy in an isobaric process

In an isobaric process, $Q=\Delta H$. So $Q = 7794.375J$

Answer:

The enthalpy change $\Delta H = 7794.375J$ and the heat $Q = 7794.375J$