Question

What is the pressure exerted by 2.50 moles of helium gas contained in a 15.0-liter container at 25.0 °C?

2.92 atm

0.342 atm

0.245 atm

4.08 atm

Answer 1

To find the pressure exerted by the helium gas, we can use the Ideal Gas Law, which is given by the formula:

\[ PV = nRT \]

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

First, we need to convert the temperature from Celsius to Kelvin:

\[
T(K) = 25.0 °C + 273.15 = 298.15 \, K
\]

Now we can plug in the values into the Ideal Gas Law and solve for pressure \( P \):

\[
n = 2.50 \, \text{moles}
\]
\[
V = 15.0 \, \text{L}
\]
\[
R = 0.0821 \, \text{L·atm/(K·mol)}
\]
\[
T = 298.15 \, K
\]

Rearranging the ideal gas law to solve for \( P \):

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

Substituting the values:

\[
P = \frac{(2.50 \, \text{mol}) \times (0.0821 \, \text{L·atm/(K·mol)}) \times (298.15 \, K)}{15.0 \, \text{L}}
\]

Calculating the numerator:

\[
2.50 \times 0.0821 \times 298.15 \approx 61.324375 \, \text{L·atm}
\]

Now divide by the volume:

\[
P \approx \frac{61.324375}{15.0} \approx 4.088 \text{ atm}
\]

Rounding to three significant figures, we get:

\[
P \approx 4.08 \, \text{atm}
\]

Thus, the pressure exerted by the helium gas is **4.08 atm**.