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What is the pressure in a 6.00 L tank with 14.4 grams of nitrogen gas at 385 K?

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Final answer:

The pressure in a 6.00 L tank with 14.4 grams of nitrogen gas at 385 K is 2.7856 atm, determined by using the ideal gas law PV = nRT.

Step-by-step explanation:

To find the pressure in a 6.00 L tank with 14.4 grams of nitrogen gas at 385 K, we use the ideal gas law which is PV = nRT. First, we need to calculate the number of moles of nitrogen gas (N₂).

The molecular weight of N₂ is approximately 28 g/mol, so 14.4 grams is 14.4 g / 28 g/mol = 0.5143 moles of nitrogen. Next, we use the ideal gas constant R = 0.0821 L·atm/(mol·K) and insert the values into the equation.

PV = nRT
P = (nRT)/V
P = (0.5143 moles * 0.0821 L·atm/(mol·K) * 385 K) / 6.00 L
P = 2.7856 atm

Therefore, the pressure in the tank is 2.7856 atm.

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