Final answer:
Using the ideal gas law PV = nRT, the temperature of a 3.0 mole gas at a pressure of 250 kPa and volume of 15 L is calculated to be approximately 241 K, with the closest answer choice being 250 K (Option C).
Step-by-step explanation:
To find the temperature of 3.0 moles of gas at a pressure of 250 kPa held in a volume of 15 L, we can use the ideal gas law which is expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin. By rearranging the formula to solve for T, we get T = PV/(nR).
Given the values:
P = 250 kPa = 250000 Pa (since 1 kPa = 1000 Pa)
V = 15 L = 0.015 m³ (since 1 L = 0.001 m³)
n = 3.0 moles
R = 8.314 J/(mol·K)
Let's plug in the values:
T = (250000 Pa × 0.015 m³) / (3.0 mol × 8.314 J/(mol·K))
T = 1500000 Pa·m³ / (24.942 J/K)
T = 60157.61 J / 24.942 J/K
= 2411.14 K
Therefore, the temperature of the gas is approximately 241 K, and the closest answer choice is Option C, 250 K.