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.