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At what temperature will 120 grams of argon gas have a volume of 0.25 Lat a pressure of

200 kPa?

1 Answer

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Answer:

-245.5°C or 27.7 K

Step-by-step explanation:

To solve this problem, we can use the ideal gas law:

PV = nRT

∴ P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we need to calculate the number of moles of argon gas using its mass and molar mass:

n = m/M

∴ m is the mass of argon and M is its molar mass. The molar mass of argon is approximately 39.95 g/mol.

n = 120 g / 39.95 g/mol = 3.003 mol

Next, we can rearrange the ideal gas law to solve for the temperature:

T = PV/nR

Substituting the given values:

T = (200 kPa) (0.25 L) / (3.003 mol) (8.314 J/(mol·K))

T = 27.7 K

Therefore, the temperature at which 120 grams of argon gas will have a volume of 0.25 L at a pressure of 200 kPa is approximately -245.5°C (27.7 K - 273.15).

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