Certainly! Let's go through the step-by-step calculations for the sum of squared errors (SSE) for both students' data.

Constants

• $R = 8.3 \, \text{J/(K·mol)}$
• $V = 16.6 \, \text{m}^3/\text{mol}$
• $n = 1 \, \text{mol}$

Ideal Gas Law

$P = \frac{nRT}{V}$

Data Collected by Student A

$T(K)$	274	276	278	282	290
$P(Pa)$	137	139	142	141	142

Data Collected by Student B

$T(K)$	276	280	284	288	290
$P(Pa)$	137	141	142	148	145

Predicted Pressures Using Ideal Gas Law

For each temperature $T$:

$P = \frac{(1 \text{ mol}) \cdot (8.3 \text{ J/(K·mol)}) \cdot T}{16.6 \text{ m}^3/\text{mol}} = \frac{8.3T}{16.6} = 0.5T$

Predicted Pressures for Student A

• $P_{A1} = 0.5 \times 274 = 137$
• $P_{A2} = 0.5 \times 276 = 138$
• $P_{A3} = 0.5 \times 278 = 139$
• $P_{A4} = 0.5 \times 282 = 141$
• $P_{A5} = 0.5 \times 290 = 145$

Predicted Pressures for Student B

• $P_{B1} = 0.5 \times 276 = 138$
• $P_{B2} = 0.5 \times 280 = 140$
• $P_{B3} = 0.5 \times 284 = 142$
• $P_{B4} = 0.5 \times 288 = 144$
• $P_{B5} = 0.5 \times 290 = 145$

Calculate SSE for Student A

\text{SSE}_A &= (137 - 137)^2 + (139 - 138)^2 + (142 - 139)^2 + (141 - 141)^2 + (142 - 145)^2 \\ &= 0 + 1 + 9 + 0 + 9 \\ &= 19 \end{aligned}$$ ### Calculate SSE for Student B $$\begin{aligned} \text{SSE}_B &= (137 - 138)^2 + (141 - 140)^2 + (142 - 142)^2 + (148 - 144)^2 + (145 - 145)^2 \\ &= 1 + 1 + 0 + 16 + 0 \\ &= 18 \end{aligned}$$ So, the SSE for Student A is 19, and for Student B, it is 18. Would you like more details or have any other questions? Here are 5 questions you can ask next: 1. Can you explain more about how to interpret SSE? 2. What if the gas constant $$R$$ was different; how would that affect the calculations? 3. Are there any other methods to measure the goodness of fit besides SSE? 4. How does temperature affect the pressure in the ideal gas law? 5. Can you provide examples of when the ideal gas law might not be accurate? Tip: Always check the units and constants used in your calculations to ensure accuracy and consistency.