Question

A 2-kg mass of helium is maintained at 300 kPa and 27 degree C in a rigid container. How large is the container, in m³?

a. A rigid tank whose volume is unknown is divided into two parts by a partition.
b. One side of the tank contains an ideal gas at 927 degree
c. The other side is evacuated and has a volume twice the size of the part containing the gas.
d. The partition is now applied and the gas expands to fill the entire tank.
e. Heat is now applied to the gas until the pressure equals the initial pressure.

Answer 1

The volume of the container holding 2 kg of helium gas at 300 kPa and 27 degrees Celsius is 41.69 m³, calculated using the Ideal Gas Law.

Step-by-step explanation:

To calculate the volume of a container holding 2 kg of helium gas at 300 kPa and 27 degrees Celsius, we use the Ideal Gas Law, which is PV = nRT. First, we need to convert the mass of helium to moles (n) using the molar mass of helium (4.00 g/mol). Since we have 2 kg of helium, this equates to 2000 g, which gives us 500 moles of helium.

Next, we convert the temperature from Celsius to Kelvin by adding 273.15, resulting in 300.15 K. The pressure (P) is already given in kilopascals, which is compatible with the units for R, the ideal gas constant (8.314 J/mol*K). Applying the Ideal Gas Law, we solve for V (volume).

V = (nRT)/P = (500 moles * 8.314 J/mol*K * 300.15 K) / (300 kPa) = 12507000 J / 300000 Pa = 41.69 m³

To determine the size of the container, we can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvin. In this case, the gas is helium and we are given P = 300 kPa, V is unknown, n = 2 kg / (4 g/mol) = 500 mol, R = 8.314 J/(mol·K), and T = 300°C + 273.15 = 573.15 K. Rearranging the equation to solve for V, we have V = (nRT) / P.

Substituting the given values, V = (500 mol)(8.314 J/(mol·K))(573.15 K) / 300 kPa = 783.9 liters = 0.7839 m³.

Therefore, the volume of the container is 41.69 m³.