Okay, let's go through this step-by-step:
1) Derive a relationship between volume and temperature using the ideal gas law:
The ideal gas law states:
PV = nRT
Where:
P = Pressure
V = Volume
n = number of moles
R = ideal gas constant
T = Temperature in Kelvin
Rearranging:
V = nRT/P
Taking the derivative with respect to T:
dV/dT = nR/P
This shows that volume is directly proportional to temperature for an ideal gas at constant pressure and amount.
2) Calculate the temperature change:
Given:
Initial Volume (V1) = 291 mL = 0.291 L
Initial Pressure (P1) = 802 torr = 1.063 atm (1 atm = 760 torr)
Initial Temperature (T1) = 28.0°C = 301.15 K
Vapor Pressure of Water at 28.0°C (Pv) = 0.0372 atm
Using ideal gas law:
P1V1 = nRT1
(1.063 atm)(0.291 L) = n(0.0821 L*atm/mol*K)(301.15 K)
n = 0.0116 mol
Partial Pressure of dry gas (Pgas) = P1 - Pv
= 1.063 atm - 0.0372 atm
= 1.026 atm
At STP:
Pressure (P2) = 1 atm
Temperature (T2) = 273.15 K
Using ideal gas law:
PV = nRT
(1 atm)(V2) = (0.0116 mol)(0.0821 L*atm/mol*K)(273.15 K)
V2 = 0.269 L
Therefore, the temperature change is:
ΔT = T2 - T1
= 273.15 K - 301.15 K
= -28 K
So the temperature decreases by 28 K when going from initial conditions to STP.