Question

1. use the appropriate simple gas law to complete the table.

missed this? read section 6.3; watch kcv 6.3
p₁(atm) 1.5 1.1 0.95
v₁(l) 7.4 6.4 17.8
n₁(mol) 0.55 0.25 0.75
t₁(k) 245 295 275
p₂(atm) 2.8 1.1 0.95
v₂(l)

n₂(mol) 0.55 0.25 1.62
t₂(k) 245 345 275

1. use the appropriate simple gas law to complete the table.

Explanation:

Step1: Identify the Gas Law

For each row, check which variables are constant. The ideal gas law is \( PV = nRT \), and for simple gas laws, we use forms like Boyle's (\( P_1V_1 = P_2V_2 \), \( n,T \) constant), Charles' (\( \frac{V_1}{T_1}=\frac{V_2}{T_2} \), \( n,P \) constant), Avogadro's (\( \frac{V_1}{n_1}=\frac{V_2}{n_2} \), \( P,T \) constant), or Combined (\( \frac{P_1V_1}{n_1T_1}=\frac{P_2V_2}{n_2T_2} \)).

Row 1 (First Row):

- \( n_1 = n_2 = 0.55 \) mol, \( T_1 = T_2 = 245 \) K (constant \( n,T \)). So use Boyle's Law: \( P_1V_1 = P_2V_2 \).
- Solve for \( V_2 \): \( V_2=\frac{P_1V_1}{P_2}=\frac{1.5 \, \text{atm} \times 7.4 \, \text{L}}{2.8 \, \text{atm}} \approx 3.99 \, \text{L} \) (Wait, but table has \( V_2 \)? Wait, no, maybe I misread. Wait the table columns: \( P_1, V_1, n_1, T_1, P_2, V_2, n_2, T_2 \). Wait first row: \( P_1=1.5 \), \( V_1=7.4 \), \( n_1=0.55 \), \( T_1=245 \), \( P_2=2.8 \), \( n_2=0.55 \), \( T_2=245 \). So \( n,T \) constant: Boyle's Law. \( V_2=\frac{P_1V_1}{P_2}=\frac{1.5\times7.4}{2.8}\approx 3.99 \approx 4.0 \) L? Wait but maybe the table is missing \( V_2 \) here? Wait no, the user's table: first row \( V_2 \) is empty? Wait the image shows first row: \( P_1=1.5 \), \( V_1=7.4 \), \( n_1=0.55 \), \( T_1=245 \), \( P_2=2.8 \), \( V_2=\)?, \( n_2=0.55 \), \( T_2=245 \). So \( V_2=\frac{1.5\times7.4}{2.8}\approx 3.99 \approx 4.0 \) L.

Row 2 (Second Row):

- \( P_1=1.1 \), \( V_1=\)?, \( n_1=0.25 \), \( T_1=295 \), \( P_2=1.1 \) atm (same \( P \)), \( V_2=6.4 \) L, \( n_2=0.25 \) mol (same \( n \)), \( T_2=345 \) K. Wait, \( P \) constant (\( P_1=P_2=1.1 \)), \( n \) constant (\( n_1=n_2=0.25 \)). So Charles' Law: \( \frac{V_1}{T_1}=\frac{V_2}{T_2} \).
- Solve for \( V_1 \): \( V_1=\frac{V_2T_1}{T_2}=\frac{6.4 \, \text{L} \times 295 \, \text{K}}{345 \, \text{K}} \approx 5.52 \, \text{L} \)? Wait no, table has \( V_1 \) as empty? Wait second row: \( P_1=1.1 \), \( V_1=\)?, \( n_1=0.25 \), \( T_1=295 \), \( P_2=1.1 \), \( V_2=6.4 \), \( n_2=0.25 \), \( T_2=345 \). So \( P,n \) constant: Charles' Law. \( V_1=\frac{V_2T_1}{T_2}=\frac{6.4\times295}{345}\approx 5.5 \) L.

Row 3 (Third Row):

- \( P_1 = P_2 = 0.95 \) atm, \( T_1 = T_2 = 275 \) K (constant \( P,T \)). So Avogadro's Law: \( \frac{V_1}{n_1}=\frac{V_2}{n_2} \).
- Solve for \( V_2 \): \( V_2=\frac{V_1n_2}{n_1}=\frac{17.8 \, \text{L} \times 1.62 \, \text{mol}}{0.75 \, \text{mol}} \approx 39.3 \, \text{L} \)? Wait table has \( V_2 \) empty? Wait third row: \( P_1=0.95 \), \( V_1=17.8 \), \( n_1=0.75 \), \( T_1=275 \), \( P_2=0.95 \), \( n_2=1.62 \), \( T_2=275 \). So \( P,T \) constant: Avogadro's Law. \( V_2=\frac{V_1n_2}{n_1}=\frac{17.8\times1.62}{0.75}\approx 39.3 \) L.

Wait, maybe the table is missing values in \( V_2 \) (row1), \( V_1 \) (row2), \( V_2 \) (row3). Let's re-express each row:

Row 1:

- Constants: \( n \) (0.55), \( T \) (245). Boyle's Law: \( P_1V_1 = P_2V_2 \)
- \( V_2 = \frac{1.5 \times 7.4}{2.8} \approx 3.99 \approx 4.0 \) L

Row 2:

- Constants: \( P \) (1.1), \( n \) (0.25). Charles' Law: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
- \( V_1 = \frac{6.4 \times 295}{345} \approx 5.5 \) L

Row 3:

- Constants: \( P \) (0.95), \( T \) (275). Avogadro's Law: \( \frac{V_1}{n_1} = \frac{V_2}{n_2} \)
- \( V_2 = \frac{17.8 \times 1.62}{0.75} \approx 39.3 \) L

But maybe the problem is to fill the missing \( V_2 \), \( V_1 \), \( V_2 \) for each row. Let's confirm the gas laws:

1. Row 1: \( n \) and \( T \) constant → Boyle's Law (\( P_1V_1 = P_2V_2 \))
2. Row 2: \( P \) and \( n \) constant → Charles' Law (\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \))
3. Row 3: \( P \) and \( T \) constant → Avogadro's Law (\( \frac{V_1}{n_1} = \frac{V_2}{n_2} \))

Now, let's calculate each missing value:

Row 1 (Missing \( V_2 \)):

\( V_2 = \frac{P_1 V_1}{P_2} = \frac{1.5 \, \text{atm} \times 7.4 \, \text{L}}{2.8 \, \text{atm}} = \frac{11.1}{2.8} \approx 3.96 \, \text{L} \approx 4.0 \, \text{L} \)

Row 2 (Missing \( V_1 \)):

\( V_1 = \frac{V_2 T_1}{T_2} = \frac{6.4 \, \text{L} \times 295 \, \text{K}}{345 \, \text{K}} = \frac{1888}{345} \approx 5.47 \, \text{L} \approx 5.5 \, \text{L} \)

Row 3 (Missing \( V_2 \)):

\( V_2 = \frac{V_1 n_2}{n_1} = \frac{17.8 \, \text{L} \times 1.62 \, \text{mol}}{0.75 \, \text{mol}} = \frac{28.836}{0.75} \approx 38.45 \, \text{L} \approx 38.5 \, \text{L} \) (Wait, 17.8×1.62=28.836; 28.836/0.75=38.448≈38.4 L)

Wait, maybe I made a miscalculation. Let's recalculate Row 3:

\( 17.8 \times 1.62 = 17.8 \times 1.6 + 17.8 \times 0.02 = 28.48 + 0.356 = 28.836 \)

\( 28.836 \div 0.75 = 38.448 \approx 38.4 \) L.

Now, let's present the calculations for each row:

Row 1 (Boyle's Law):

- Given: \( P_1 = 1.5 \) atm, \( V_1 = 7.4 \) L, \( P_2 = 2.8 \) atm, \( n_1 = n_2 = 0.55 \) mol, \( T_1 = T_2 = 245 \) K.
- Formula: \( P_1V_1 = P_2V_2 \)
- \( V_2 = \frac{P_1V_1}{P_2} = \frac{1.5 \times 7.4}{2.8} \approx 3.96 \) L (≈4.0 L)

Row 2 (Charles' Law):

- Given: \( V_2 = 6.4 \) L, \( T_1 = 295 \) K, \( T_2 = 345 \) K, \( P_1 = P_2 = 1.1 \) atm, \( n_1 = n_2 = 0.25 \) mol.
- Formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
- \( V_1 = \frac{V_2 T_1}{T_2} = \frac{6.4 \times 295}{345} \approx 5.47 \) L (≈5.5 L)

Row 3 (Avogadro's Law):

- Given: \( V_1 = 17.8 \) L, \( n_1 = 0.75 \) mol, \( n_2 = 1.62 \) mol, \( P_1 = P_2 = 0.95 \) atm, \( T_1 = T_2 = 275 \) K.
- Formula: \( \frac{V_1}{n_1} = \frac{V_2}{n_2} \)
- \( V_2 = \frac{V_1 n_2}{n_1} = \frac{17.8 \times 1.62}{0.75} \approx 38.45 \) L (≈38.5 L)

Assuming the table's missing values are \( V_2 \) (row1), \( V_1 \) (row2), \( V_2 \) (row3), these are the calculations.

Answer:

Row 1 \( V_2 \): \( \approx 4.0 \) L

Row 2 \( V_1 \): \( \approx 5.5 \) L

Row 3 \( V_2 \): \( \approx 38.4 \) L

(Note: Depending on rounding, values may vary slightly. For example, Row 1: \( 1.5×7.4=11.1 \); \( 11.1÷2.8≈3.96 \) L (≈4.0 L). Row 2: \( 6.4×295=1888 \); \( 1888÷345≈5.47 \) L (≈5.5 L). Row 3: \( 17.8×1.62=28.836 \); \( 28.836÷0.75=38.448 \) L (≈38.4 L).)