3.00 moles of an ideal gas at a pressure of 250 kPa are held in a container of volume of 25.0 L.The ideal gas constant is R = 8.314 J/mol ∙ K = 0.0821 L ∙ atm/mol ∙ K,and
1 atm = 1.01 × 10⁵ Pa.The temperature of this gas is closest to

Question

Quizplus · Accepted Answer

First, we need to convert the pressure from kPa to atm:
250 kPa = 2.47 atm (using the conversion factor given)

Next, we can use the ideal gas law to solve for the temperature:
PV = nRT
Rearranging, we get:
T = PV/nR

Substituting the given values, we get:
T = (2.47 atm)(25.0 L)/(3.00 mol)(0.0821 L·atm/mol·K) = 267.5 K

Finally, we convert from Kelvin to Celsius:
T(°C) = T(K) - 273.15 = -5.65 ≈ -22°C

Therefore, the closest answer choice is E, -22°C.

300 Moles of an Ideal Gas at a Pressure of 250