Answer :
The pressure inside the container is 1761.44 kPa.
Determine the pressure
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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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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