Final answer:
The temperature of 4.45 g of helium gas at a pressure of 1.70 atm and a volume of 16.3 L is 329.14 K
Step-by-step explanation:
The question asks for the temperature of 4.45 g of helium gas at a pressure of 1.70 atm and a volume of 16.3 L. To find the temperature in kelvins, we can use the Ideal Gas Law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in kelvins. Rearranging the equation to solve for T, we have T = PV / nR. To use this equation, we need to calculate the number of moles of helium gas using its molar mass. The molar mass of helium is 4.0026 g/mol, so the number of moles is n = 4.45 g / 4.0026 g/mol. We can substitute this value along with the given values of pressure, volume, and the ideal gas constant (R = 0.0821 L·atm/mol·K) into the equation to find the temperature in kelvins.
Using the given values:
- Pressure (P) = 1.70 atm
- Volume (V) = 16.3 L
- Molar mass of helium (M) = 4.0026 g/mol
- Ideal gas constant (R) = 0.0821 L·atm/mol·K
Calculating the number of moles:
n = 4.45 g / 4.0026 g/mol = 1.1115 mol
Substituting the values into the equation for temperature:
T = (1.70 atm)(16.3 L) / (1.1115 mol)(0.0821 L·atm/mol·K) = 329.14 K
Therefore, the temperature of 4.45 g of helium gas at a pressure of 1.70 atm and a volume of 16.3 L is 329.14 K, rounded to three significant figures.