Final answer:
To calculate the temperature of the ideal gas, we apply the Ideal Gas Law, which gives 238 Kelvin as the temperature of the 2.51 mol gas sample at 1.60 atm and 30.67 L.
Step-by-step explanation:
To find the temperature of a sample of an ideal gas given its pressure, volume, and amount in moles, we can use the Ideal Gas Law, which is PV = nRT, where 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 convert the given pressure from atm to the standard unit of pressure in the Ideal Gas Law, which is Pascals (Pa). However, for the purposes of this exercise, we can use the R constant value which has the atm unit, so the conversion is not necessary. Taking R = 0.0821 L·atm/K·mol (which is a common value for R when pressure is in atm, volume in liters and temperature in Kelvin), we can rearrange the Ideal Gas Law to solve for T:
T = (PV) / (nR)
Plugging in the given values:
T = (1.60 atm × 30.67 L) / (2.51 mol × 0.0821 L·atm/K·mol)
Calculating T:
T = 49.072 / 0.206151 = 238 K
Therefore, the temperature of the sample is 238 Kelvin.