Determine the volume occupied by 2kg of an ideal gas at 120kpa and 300k. Assume suitable value of the gass?

To determine the volume occupied by 2kg of an ideal gas, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in pascals)
V = volume (in cubic meters)
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)

First, we need to convert the given pressure from kilopascals to pascals:
120 kPa = 120,000 Pa

Next, we need to convert the given temperature from degrees Celsius to Kelvin:
300 K = 300 + 273.15 = 573.15 K

Now, we can rearrange the ideal gas law equation to solve for volume (V):
V = (nRT) / P

To find the number of moles (n) of gas, we can use the equation:
n = mass / molar mass

Assuming a suitable value for the gas, let's use the molar mass of helium (He), which is approximately 4 g/mol.

Converting the given mass of 2 kg to grams:
2 kg = 2,000 g

Now, we can calculate the number of moles (n):
n = 2,000 g / 4 g/mol = 500 mol

Plugging in the values into the volume equation:
V = (500 mol * 8.314 J/(mol·K) * 573.15 K) / 120,000 Pa

Calculating the volume:
V = 19.82 m³

Therefore, 2 kg of the ideal gas occupies a volume of approximately 19.82 cubic meters at 120 kPa and 300 K.