Answer :
At a pressure of 1.13 atm and a volume of 35.9 L, 0.750 mol of an ideal gas will occupy a temperature of approximately 387.66°C.
Given information,
Pressure (P) = 1.13 atm
Volume (V) = 35.9 L
Number of moles (n) = 0.750 mol
The ideal gas law equation: PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Now,
T = PV / (nR)
T = (1.13 * 35.9 ) / (0.750 * 0.0821)
T = (40.607 atm·L) / (0.061575 mol·L/(K·atm))
T ≈ 660.81 K
T(°C) = T(K) - 273.15
T(°C) ≈ 660.81 - 273.15
T(°C) ≈ 387.66°C
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