. Calculate the approximate volume of a 1.50 mol sample of gas at 300K and a pressure of 3.75 atm.
4. Pressure CO2 = 760 mm Hg Volume CO2 = 2.965 L Temperature CO2= 25.5 °C
How many moles of CO2 were produced?
5. What instrument do you measure pressure with?
6. The volume of a gas is 500 mL at 20.0 K. What will be the new temperature if the gas is compressed to 100 mL?
. The volume of a sample of oxygen is 200.0 mL when the pressure is 4.0 atm and the temperature is 37.0 °C. What is the new temperature if the volume increases to 400.0 mL and the pressure decreases to 2.000 atm?
8. The pressure on a 2.00L sample of CO2(g) is increased from 50 kPa to 100 kPa. What is the new volume of the gas?
9. The temperature of a 2.0 liter sample of helium gas at STP is increased to 37°C, and the pressure is decreased to 80 kPa. What is the new volume of the helium sample?

Respuesta :

Answer:

3. 9.84 L.

4. 0.1211 mol.

5. Barometer, Bourdon tube, and Aneroid barometers.

6. 4.0 K.

7. 310 K.

8. 1.0 L.

9. 2.637 L.

Explanation:

3. Calculate the approximate volume of a 1.50 mol sample of gas at 300K and a pressure of 3.75 atm.

  • To calculate the no. of moles of a gas, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 3.75 atm).

V is the volume of the gas in L.

n is the no. of moles of the gas in mol (n = 1.5 mol).

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K (T = 300 K).

∴ V = nRT/P = (1.5 mol)( 0.082 L.atm/mol.K)( 300 K)/(3.75 atm) = 9.84 L.

4. Pressure CO2 = 760 mm Hg Volume CO2 = 2.965 L Temperature CO2= 25.5 °C . How many moles of CO2 were produced?

∵ P = 760 mmHg = 1.0 atm, V = 2.965 L, T = 25.5°C + 273 = 298.5 K, R = 0.082 L.atm/mol.K.

∴ n = PV/RT = (1.0 atm)(2.965 L)/(0.082 L.atm/mol.K)(298.5 K) = 0.1211 mol.

5. What instrument do you measure pressure with?

Barometer, Bourdon tube, and Aneroid barometers.

  • A barometer is a scientific instrument used to measure air pressure. Pressure tendency can forecast short term changes in the weather. Many measurements of air pressure are used within surface weather analysis to help find surface troughs, high pressure systems and frontal boundaries.
  • Another type is the Bourdon tube. It consists of a flattened tube that is shaped like a “C.” When the pressure outside the tube is greater than the pressure inside, the “C” curls up more. When the pressure inside is greater the “C” uncurls more.  
  • A third type is Aneroid barometers. They use vacuum capsules called cells. As air pressure changes, the cell expands and contracts, and a device measure this.

6. The volume of a gas is 500 mL at 20.0 K. What will be the new temperature if the gas is compressed to 100 mL?

If n and P are constant, and have two different values of V and T:

V₁T₂ = V₂T₁

V₁ = 500 mL, T₁ = 20 K.

V₂ = 100 mL, T₂ =??? K.

∴ T₂ = V₂T₁/V₁ = (100 mL)(20 K)/(500 mL) = 4.0 K.

7. The volume of a sample of oxygen is 200.0 mL when the pressure is 4.0 atm and the temperature is 37.0 °C. What is the new temperature if the volume increases to 400.0 mL and the pressure decreases to 2.000 atm?

  • If n is constant, and have two different values of (P, V and T):

P₁V₁T₂ = P₂V₂T₁

P₁ = 4.0 atm, V₁ = 200 mL, T₁ = 37°C + 273 = 310 K.

P₂ = 2.0 atm, V₂ = 400 mL, T₂ =??? K.

∴ T₂ = P₂V₂T₁/P₁V₁ = (2.0 atm)(400 mL)(310 K)/(4.0 atm)(200 mL) = 310 K.

8. The pressure on a 2.00L sample of CO2(g) is increased from 50 kPa to 100 kPa. What is the new volume of the gas?

  • If n and T are constant, and have two different values of V and P:

P₁V₁ = P₂V₂

P1 = 50 kPa, V1 = 2.0 L.

P₂ = 100 kPa, V₂ = ??? L.

∴ V₂ = P₁V₁/P₂ = (50 kPa)(2.0 L)/(100 kPa) = 1.0 L.

9. The temperature of a 2.0 liter sample of helium gas at STP is increased to 37°C, and the pressure is decreased to 80 kPa. What is the new volume of the helium sample?

  • If n is constant, and have two different values of (P, V and T):

P₁V₁T₂ = P₂V₂T₁

P₁ = 1.0 atm (standard P), V₁ = 2.0 L, T₁ = 25°C + 273 = 298 K (standard T).

P₂ = 80 kPa = 0.789 atm, V₂ = ??? L, T₂ = 37°C + 273 = 310 K.

V₂ = P₁V₁T₂/P₂T₁ = (1.0 atm)(2.0 L)(310 K)/(0.789 atm)(298 K) = 2.637 L.