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Determine the temperature (in K) of 2.14 moles of gas contained in a 3.09 L vessel at a pressure of 142.01 kPa.

User Woodsy
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Answer:

We can use the ideal gas law to solve for the temperature of the gas:

PV = nRT

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

R = 8.31 J/(mol K) is the ideal gas constant.

First, we need to convert the pressure from kPa to Pa:

P = 142.01 kPa = 142,010 Pa

Next, we can solve for T:

T = PV/nR

T = (142010 Pa)(3.09 L)/(2.14 mol)(8.31 J/(mol K))

T = 229 K

Therefore, the temperature of the gas is 229 K.

User John Taylor
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