According t o the ideal gas law, a 0 . 9 1 1 5 mol sample o f krypton gas i n a 1 . 1 6 6 L container a t 2 6 5 . 1 K should exert a pressure o f 1 7 . 0 1 atm . What i s the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure?
Question:
According t o the ideal gas law, a 0 . 9 1 1 5 mol sample o f krypton gas i n a 1 . 1 6 6 L container a t 2 6 5 . 1 K should exert a pressure o f 1 7 . 0 1 atm . What i s the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure?
This is AI generated Response.
Asked by: Test User 100111
Created at: 2025-05-07 06:55:07
Asked by: Test User 100111
Created at: 2025-05-07 06:55:07
Final Answer:
Rendered
Raw Text
Did you like this answer?
Let's walk through the calculation step by step:
### Given Data
- **n** (moles of Kr) = 0.9115 mol
- **V** (volume) = 1.166 L
- **T** (temperature) = 265.1 K
- **\( P_{ideal} \)** (ideal pressure) = 17.01 atm
### Constants for Krypton (Kr)
From standard tables:
- **a** = 2.318 L²·atm·mol⁻²
- **b** = 0.0398 L·mol⁻¹
- **R** = 0.08206 L·atm·mol⁻¹·K⁻¹
---
## 1. **Calculate Pressure with the van der Waals Equation**
The van der Waals equation is:
\[
\left( P + a \left( \frac{n}{V} \right)^2 \right)(V - n b) = n R T
\]
Solve for \( P \):
\[
P = \frac{nRT}{V - nb} - a \left(\frac{n}{V}\right)^2
\]
#### **Calculate each term:**
1. \( nb = 0.9115 \times 0.0398 = 0.03629 \) L
2. \( V - nb = 1.166 - 0.03629 = 1.12971 \) L
3. \( nRT = 0.9115 \times 0.08206 \times 265.1 = 19.849 \)
4. \( \frac{nRT}{V - nb} = \frac{19.849}{1.12971} = 17.57 \) atm
Now, \( \left( \frac{n}{V} \right)^2 = \left( \frac{0.9115}{1.166} \right)^2 = (0.7817)^2 = 0.6111 \)
So,
\[
a \left(\frac{n}{V}\right)^2 = 2.318 \times 0.6111 = 1.417 \text{ atm}
\]
#### **Plug these into the equation:**
\[
P_{vdW} = 17.57 - 1.417 = 16.15 \text{ atm}
\]
---
## 2. **Percent Difference**
\[
\text{Percent Difference} = \frac{|P_{ideal} - P_{vdW}|}{P_{ideal}} \times 100\%
\]
\[
= \frac{|17.01 - 16.15|}{17.01} \times 100\%
\]
\[
= \frac{0.86}{17.01} \times 100\%
\]
\[
= 5.06\%
\]
---
## **Final Answer**
**The percent difference between the van der Waals pressure and the ideal pressure is:**
\[
\boxed{5.06\%}
\]
**(rounded to three significant figures)**
---
**Summary Table**
| Method | Pressure (atm) |
|------------------------- |---------------|
| Ideal Gas Law | 17.01 |
| van der Waals Equation | 16.15 |
| Percent Difference | 5.06% |
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :
VIPSolutions