Question

Report your answer to 3 significant figures and do not include units in your answer. Derive a relationship between volume and temperature using the ideal gas law. Note: the problem asks for the temperature change. Tries 1/99 Previous Tries wet gas collected occupies a volume of 291 mL at a pressure of 802 torr and temperature of 28. 0∘C. The vapor pressure of water at 28. 0∘C is 0. 0372 atm. 1. Calculate the volume (L) that the gas occupies after it is cried (the water vapor removed) and stored at STP. 2. 69×10−1 L The mass of the dry gas is 479mg. A fragment of the tank indicates that the gas is a monoatomic element. 2. Write out the full name of the gas. Given, the molar mas of the gas can be calculated (molar mass = mass/mol). To find the correct gas, look at the Periodic Table. Tries 19/99 Previous Trios At what temperature (in ∘C ) does 10. 89 g of helium occupy a volume of 7. 18 L at a pressure of 1. 13 atm ? Report your answer to 3 significant figures and do not include units in your answer. ∘C

Answer 1

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.