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About 2 L. Assuming an ideal gas, use the ideal gas law to calculate the volume.

PV = nRT or V = nRT/P where n = 4.24 g / 28 g/mol or 0.151 mol N2 and R = 0.08206 L*atm/mol*K and T = 331 K

V = (0.151 mol)*(0.08206 L*atm/mol*K)*(331K) / (2.04 atm)

V = 1.997 L

Great job on the answer! I would just change one thing: the atm/mol*K I would change to either atm/(mol*K) or atm/mol/K.

It's been a while since I've taken chemistry, and this helps me to help my daughter with her chemistry. Thank you for taking the time to answer this question.

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13y ago
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9y ago

The volume of 0.625 mol of nitrogen at 74.2 kPa and 85 degrees Celsius can be calculated using the formula Volume equals moles times gas constant times temperature divided by pressure. The gas constant is .08296 Latm/molK, 74.2 kPA in atm is .732, and 85 Celsius in kelvin is 358.15. The final result is 25.08 liters.

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10y ago

Use the ideal gas law equation, which is expressed as PV=nRT. P (pressure) is 6.5atm, n (moles) is 2.12, R, the ideal gas constant, is 0.0812, T (temperature) is 76 Celsius, but you must add 273 to it, so that it will be in kelvins. Now, plug it all in: 6.5(V)=2.12(0.0812)(76+273)=approx. 60.1, divided by 6.5=9.25 liters.

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9y ago

According to the ideal gas law, PV = nRT, where

P = pressure = 0.82atm

V = volume = ?

n = amount of substance in moles = 2.5mol

R = gas constant = 0.08206 L atm/mol K

T = temperature in Kelvins = 57 + 273.15 = 330.15K

Solve for V:

V = nRT/P = 2.5mol x 0.08206 L atm/mol K x 330.15K/0.82atm = 83L (82.598 L rounded to two significant figures due to 2.5mol and 0.82atm.)

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7y ago

The volume of 2 moles of nitrogen gas at room temperature and 1 atmosphere is 48,93 L.

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13y ago

6 L

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10y ago

101L

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12y ago

5.60 L

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Q: According to the ideal gas law what will be the volume of 2.5 mol of nitrogen at 0.82 atm pressure 57 C temperature?
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