What is the number of molecules in an 80 L gas with a gauge pressure of 2 atm at 300 K?

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What is the number of molecules in an 80 L gas with a gauge pressure of 2 atm at 300 K?

To determine the number of molecules in an 80 L gas at a gauge pressure of 2 atm and a temperature of 300 K, the ideal gas law can be utilized, which is expressed as: \[ PV = nRT \] In this equation: - \( P \) is the absolute pressure, - \( V \) is the volume, - \( n \) is the number of moles, - \( R \) is the ideal gas constant (approximately 0.0821 L·atm/(K·mol)), - \( T \) is the temperature in Kelvin. First, it’s important to convert the gauge pressure to absolute pressure. Since gauge pressure measures pressure above atmospheric pressure, we add the atmospheric pressure (approximately 1 atm): \[ P = 2 \, \text{atm} + 1 \, \text{atm} = 3 \, \text{atm} \] Now substituting the values into the ideal gas law gives us: \[ (3 \, \text{atm}) (80 \, \text{L}) = n (0.0821 \, \text{L·atm/(K·mol)}) (300 \, K) \] Calculating

To determine the number of molecules in an 80 L gas at a gauge pressure of 2 atm and a temperature of 300 K, the ideal gas law can be utilized, which is expressed as:

[ PV = nRT ]

In this equation:

  • ( P ) is the absolute pressure,

  • ( V ) is the volume,

  • ( n ) is the number of moles,

  • ( R ) is the ideal gas constant (approximately 0.0821 L·atm/(K·mol)),

  • ( T ) is the temperature in Kelvin.

First, it’s important to convert the gauge pressure to absolute pressure. Since gauge pressure measures pressure above atmospheric pressure, we add the atmospheric pressure (approximately 1 atm):

[ P = 2 , \text{atm} + 1 , \text{atm} = 3 , \text{atm} ]

Now substituting the values into the ideal gas law gives us:

[ (3 , \text{atm}) (80 , \text{L}) = n (0.0821 , \text{L·atm/(K·mol)}) (300 , K) ]

Calculating

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