Question

1-If 3.89 moles of an ideal gas has a pressure of 1.42 atm, and a volume of 26.99 L, what is the temperature of the sample in degrees Celsius?

2-A sample of an ideal gas has a volume of 3.40 L at 14.20 °C and 1.30 atm. What is the volume of the gas at 22.40 °C and 0.987 atm? answer____L

Answer 1

1) Use the ideal gas law.

P*V = n*R*T

where P = pressure of the ideal gas = 1.42 atm; V = volume of the ideal gas = 26.99 L and n = number of moles of the ideal gas = 3.89 moles. We are required to find out T, the temperature of the gas in the Kelvin scale.

T = PV/nR = (1.42 atm)*(26.99 L)/(3.89 mole)(0.082 L-atm/mol.K) = 120.15 K.

We know that T (K) = t(°C) + 273.15 K

===> t(°C) = 120.15 K – 273.15 K = -153.

The temperature of the ideal gas is -153°C (ans).

2) Use Boyle’s law of pressure and volume.

P₁*V₁/T₁ = P₂*V₂/T₂

where P₁ = 1.30 atm, V₁ = 3.40 L and T₁ = 14.20°C = (14.20 + 273.15) K = 287.35 K; P₂ = 0.987 atm and T₂ = 22.40°C = (22.40 + 273.15) K = 295.55 K. Plug in values and obtain V₂.

V₂ = P₁*V₁*T₂/P₂T₁ = (1.30 atm)*(3.40 L)*(295.55 K)/(0.987 atm)(287.35 K) = 4.6060 L ? 4.61 L.

The volume of the gas is 4.61 L (ans).