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What is the pressure in atm, of 6.00 grams of Hydrogen gas (H

2

​

) in 800.ml and 212. F? 115 atm 900 atm 22.4 atm 0.0821 atm

Answer :

The pressure of 6.00 grams of Hydrogen gas (H₂) in 800 mL and 212 °F is approximately 115 atm.To find the pressure of the hydrogen gas (H₂) in atm, we need to use the ideal gas law equation:

PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant (0.0821 atm L/mol K)
T = temperature in Kelvin

First, we need to convert the given volume from milliliters (ml) to liters (L). Since there are 1000 mL in 1 L, we have:
800 mL ÷ 1000 mL/L = 0.8 L

Next, we need to convert the given mass of hydrogen gas (H₂) from grams to moles. To do this, we divide the mass by the molar mass of hydrogen, which is 2 g/mol:
6.00 g ÷ 2 g/mol = 3.00 mol

Now, we have all the values we need to calculate the pressure. However, we need to convert the given temperature from Fahrenheit (°F) to Kelvin (K). The conversion formula is:
T(K) = (T(°F) - 32) * 5/9 + 273.15

Let's calculate the temperature in Kelvin:
T(K) = (212 °F - 32) * 5/9 + 273.15
T(K) = (180) * 5/9 + 273.15
T(K) = 100 + 273.15
T(K) = 373.15 K

Now, we can substitute the values into the ideal gas law equation:
P * 0.8 L = 3.00 mol * 0.0821 atm L/mol K * 373.15 K
P * 0.8 L = 92.72775 atm L

Dividing both sides by 0.8 L:
P = 92.72775 atm ÷ 0.8 L
P = 115.90968 atm
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