What is the pressure inside a 35.5 L container holding 101.5 kg of argon gas at 23.0°C?

The pressure inside a 35.5 L container holding 101.5 kg of argon gas at 23.0 °C can be found using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the mass of argon gas to moles.

The molar mass of argon is 39.95 g/mol, so: 101.5 kg * 1000 g/kg / 39.95 g/mol = 2540.68 mol

Next, we need to convert the temperature from Celsius to Kelvin: 23.0 °C + 273.15 = 296.15 K

Now we can plug in the values into the ideal gas law and solve for the pressure:

P = nRT / V

P = (2540.68 mol)(8.314 J/mol*K)(296.15 K) / (35.5 L)

P = 1761437.58 J/L P = 1761.44 kPa

Therefore, the pressure inside the container is 1761.44 kPa.

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cinnamonbearsssss cinnamonbearsssss · Answer 2 · 2024-09-04T18:54:42-04:00

The pressure inside the container is 1706.28 atm. This is calculated by using the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature.

The pressure inside a 35.5 L container holding 101.5 kg of argon gas at 23.0 °C can be calculated using the ideal gas law equation, PV = nRT. First, we need to convert the mass of argon gas to moles:

101.5 kg / 39.95 g/mol = 2540.18 mol

Next, we need to convert the temperature from Celsius to Kelvin:

23.0 °C + 273.15 = 296.15 K

Now we can plug in the values into the ideal gas law equation:

P(35.5 L) = (2540.18 mol)(0.08206 L·atm/mol·K)(296.15 K)

Solving for pressure:

P = (2540.18 mol)(0.08206 L·atm/mol·K)(296.15 K) / 35.5 L

P = 1706.28 atm

Therefore, the pressure inside the container is 1706.28 atm.

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