Dll Lab Sec 1, A sample of an ideal gas at 25 °C occupies 3.00 Lat 2.00 atm. Wha

Question

Dll Lab Sec 1, A sample of an ideal gas at 25 °C occupies 3.00 Lat 2.00 atm. What volume will it occupy at 10 and the same pressure? A. 1.20 L B, 8.33 L 3.00 L D, 2,85 L E, 3,16 L 2. Consider a container filled with a gas. Which of the following will increase the pressure in the container? 1, Add gas to the container without changing the volume or temperature. 2, Raise the temperature without changing the volume. 3. Increase the volume without changing the temperature. A. 1, 2, and 3 B. Only 3 C. 2 and 3 D. 1 and 3 E. 1 and 2 3. What is the molar mass of a gas if a sample has a density of g/L at 1.00 atm and o c? Hint: Consider a specific volume of the gas. A. 13.1 g/mol B. 22.4 g/mol C. 5.86 g/mol D. 120. g/mol E. 131 g/mol 4. mixture of gases that is 20% o2 and 80% N2 (by moles) is collected over water at 20°C and a A total pressure of 1.00 atm. The vapor pressure of water is 18 mm Hg at this temperature. Put the gases in order from the one with the lowest partial pressure to the one with the highest. Highest Lowest PH20 A. Po2 N2 B. Po2 PH20 PN2 PH20 C. PN2 D. PH20 Po2 PN2 E. PH20 N2 Po2

Explanation / Answer

1) we know that

PV=nRT

given pressure is constant

so

V2/ V1 = T2/T1

V2 = V1 T2 / T1

so

V2 = 3 x 283 / 298

V2 = 2.85 L

so the answer is option D

2) use PV = nRT

P= nRT / V

so the answer is option E. 1 and 2

3) PV = nRT

but n = m/ M

so

PV = mRT / M

consider 1 L of sample

mass = desnity x volume

mass = 5.86 x 1 = 5.86 g

so

PMV = mRT

so

1 x M x 1 = 5.86 x 0.0821 x 273

M = 131 g/mol

so the answer is option E . 131 g/mol

4)   partial pressure = mole fraction x total pressure

P02 = ( 0.2 ) x 1 = 0.2 atm

PN2 = 0.8 x 1 = 0.8 atm

PH20 = 18 / 760 = 0.0237 atm

so the correct answer is option D

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

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