Question

31. Based on Figure 2, it 13 g of Kr had been added to the
6 L vessel, the pressure would have been:
A. less than 200 torr.
B. between 200 torr and 400 torr.
C. between 400 torr and 600 torr.
D. greater than 600 torr.

Answer 1

To determine the pressure of 13 g of Kr in a 6 L vessel, you can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm or torr)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol) or 62.36 L·torr/(K·mol))
- \( T \) = temperature (in Kelvin)

First, we need to convert the mass of Kr to moles. The molar mass of krypton (Kr) is approximately 83.8 g/mol.

Calculating the number of moles (\( n \)):
\[
n = \frac{mass}{molar \: mass} = \frac{13 \: g}{83.8 \: g/mol} \approx 0.155 \: mol
\]

Next, you need to assume a temperature to calculate the pressure. If we take room temperature to be approximately 25°C, then in Kelvin this is:
\[
T = 25 + 273.15 = 298.15 \: K
\]

Now, we can calculate the pressure using the Ideal Gas Law. We'll use the gas constant in torr:
\[
R = 62.36 \: L \cdot torr/(K \cdot mol)
\]

Rearranging the Ideal Gas Law to solve for pressure \( P \):
\[
P = \frac{nRT}{V}
\]

Substituting the values we have:
\[
P = \frac{(0.155 \: mol)(62.36 \: L \cdot torr/(K \cdot mol))(298.15 \: K)}{6 \: L}
\]

Calculating:
\[
P \approx \frac{(0.155)(62.36)(298.15)}{6} \approx \frac{(0.155)(1858.05)}{6} \approx \frac{288.93}{6} \approx 48.16 \: torr
\]

Upon recalculating this with attention to exact values,
\[
P \approx 480.2 \: torr
\]

So, the pressure would be:
C. between 400 torr and 600 torr.