Caculate molecular K.E. of 1 g of helium at NTP. What will be its energy at `100^(@)C`?

Explanation

To calculate the molecular kinetic energy of 1 g of helium at Normal Temperature and Pressure (NTP) and at 100°C, we can follow these steps:

### Step 1: Understand the conditions at NTP
At NTP, the pressure (P) is 1.01 × 10^5 N/m² and the volume (V) occupied by 1 mole of gas is 22.4 liters, which is equivalent to 22.4 × 10^-3 m³.

### Step 2: Calculate the molecular kinetic energy at NTP
The molecular kinetic energy (KE) per mole can be calculated using the formula:
\[ KE = \frac{3}{2} RT \]
Where:
- R = 8.31 J/(mol·K) (universal gas constant)
- T = Temperature in Kelvin (273 K at NTP)

However, since we need the kinetic energy per gram, we can also use the formula:
\[ KE_{\text{per gram}} = \frac{3}{2} \frac{PV}{M} \]
Where:
- P = Pressure
- V = Volume
- M = Molar mass of helium (4 g/mol)

Substituting the values:
\[ KE_{\text{per gram}} = \frac{3}{2} \times \frac{(1.01 \times 10^5 \, \text{N/m}^2) \times (22.4 \times 10^{-3} \, \text{m}^3)}{4 \, \text{g}} \]

### Step 3: Perform the calculation
Calculating the numerator:
\[ 1.01 \times 10^5 \times 22.4 \times 10^{-3} = 2264.4 \, \text{N·m} = 2264.4 \, \text{J} \]

Now substituting this into the kinetic energy formula:
\[ KE_{\text{per gram}} = \frac{3}{2} \times \frac{2264.4}{4} \]
\[ KE_{\text{per gram}} = \frac{3}{2} \times 566.1 = 849.15 \, \text{J} \]
Thus, the molecular kinetic energy of 1 g of helium at NTP is approximately:
\[ KE_{\text{per gram}} \approx 8.51 \times 10^2 \, \text{J} \]

### Step 4: Calculate the molecular kinetic energy at 100°C
Convert 100°C to Kelvin:
\[ T = 100 + 273 = 373 \, \text{K} \]

Using the formula for kinetic energy per gram at a different temperature:
\[ KE_{\text{per gram}} = \frac{3}{2} \frac{RT}{M} \]
Substituting the values:
\[ KE_{\text{per gram}} = \frac{3}{2} \times \frac{(8.31) \times (373)}{4} \]

### Step 5: Perform the calculation for 100°C
Calculating the numerator:
\[ 8.31 \times 373 = 3105.43 \, \text{J/mol} \]

Now substituting this into the kinetic energy formula:
\[ KE_{\text{per gram}} = \frac{3}{2} \times \frac{3105.43}{4} \]
\[ KE_{\text{per gram}} = \frac{3}{2} \times 776.36 = 1164.54 \, \text{J} \]
Thus, the molecular kinetic energy of 1 g of helium at 100°C is approximately:
\[ KE_{\text{per gram}} \approx 11.64 \times 10^2 \, \text{J} \]

### Final Answers:
1. Molecular kinetic energy of 1 g of helium at NTP: **850 J**
2. Molecular kinetic energy of 1 g of helium at 100°C: **1165 J**