90 g glucose is filled in a container at 10 atm, if the volume of container is 2 liter, what is the temperature of glucose (z = 1.5)?

Question

Quizplus · Accepted Answer

The ideal gas law, modified by the compressibility factor (Z), is PV = ZnRT. First, calculate the number of moles of glucose (C6H12O6, molar mass = 180 g/mol). Moles of glucose = 90 g / 180 g/mol = 0.5 mol. Given: P = 10 atm, V = 2 L (converted to m^3 for consistency in units, 1 L = 0.001 m^3, so V = 0.002 m^3), Z = 1.5, R = 0.0821 L·atm/(mol·K) (or 8.314 J/(mol·K) in SI units). To find T in Kelvin, rearrange the equation to T = PV/(ZnR). Using SI units for R, T = (10 atm * 0.002 m^3) / (1.5 * 0.5 mol * 8.314 J/(mol·K)). Convert atm to Pa for consistency, 1 atm = 101325 Pa, so P = 10 * 101325 Pa. T = (10 * 101325 * 0.002) / (1.5 * 0.5 * 8.314) = 324.8 K. Convert to °C: T = 324.8 K - 273.15 = 51.65°C, which rounds to 51.7°C.

90 G Glucose Is Filled in a Container at 10