Pressure-Temperature Relationship in Gases Advanced Study Name Lab Section 1. A

Question

Pressure-Temperature Relationship in Gases Advanced Study Name Lab Section 1. A flask having a volume of 134 mL is heated in a boiling water at a temperature of 99.8C and a pressure of 753 mm Hg. The flask is then cooled to 2.0°C in an ice-water bath. a. Determine the Kelvin temperature of the boiling water bath b. Determine the Kelvin temperature of the ice-water bath c. Determine the pressure inside the flask at 2.0°C mm Hg d. Express this pressure in atmospheres and kilopascals atm kPa 2. If the same flask was cooled down to a temperature of 2 Kelvin, what would the pressure inside the flask be in mm Hg? mm Hg 3. A gas in a sealed container with a volume of 4.35 L is heated from 150°C to 450°C. If the initial pressure was 2.4 atm, what is the final pressure in atm and in mm Hg? atm _ mm Hg

Explanation / Answer

1)

a) 99.8°C + 273.15 = 372.95K

b) 2°C + 273.15K = 275.15K

c) According - Lussac law

P = C T

P1/T1 = P2/T2

P2 = T2(P1/T1)

= 2°C(753mmHg/99.8°C)

= 15.09mmHg

d) 760mmHg = 1atm

(1atm/760mmHg)*15.09mmHg = 0.0198atm

1mmHg = 0.13332239

15.09mmHg * 0.13332239 = 2.0118kPa

2) P2 = T2(P1/T1)

= 2K ( 753mmHg /372.95K)

= 4.04mmHg

3) P2 = T2 (P1/T1)

= 450°C (2.4atm/150°C)

= 7.2atm

1atm = 760mmHg

7.2atm = 7.2 * 760mmHg =5472 mmHg

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.