Question

What pressure, in atm, is exerted by 2.50 L of gas containing 6.35 mol at 320 K?
a
14.19 atm
b
22.54 atm
c
18.42 atm
d
66.7 atm

Answer 1

To calculate the pressure exerted by a gas using the ideal gas law, we can use the formula:

\[
PV = nRT
\]

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

Given:
- \( V = 2.50 \, \text{L} \)
- \( n = 6.35 \, \text{mol} \)
- \( T = 320 \, \text{K} \)

We need to solve for \( P \):

\[
P = \frac{nRT}{V}
\]

Now we can plug in the values:

\[
P = \frac{(6.35 \, \text{mol}) \cdot (0.0821 \, \text{L·atm/(K·mol)}) \cdot (320 \, \text{K})}{2.50 \, \text{L}}
\]

Calculating the numerator:

\[
6.35 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 320 \, \text{K} = 167.1052 \, \text{L·atm}
\]

Now divide by the volume:

\[
P = \frac{167.1052 \, \text{L·atm}}{2.50 \, \text{L}} = 66.84208 \, \text{atm}
\]

Rounding to three significant figures, the pressure \( P \) is approximately:

\[
P \approx 66.8 \, \text{atm}
\]

Thus, the option that matches this result is:

**d) 66.7 atm** (after rounding).