Question

If an 85.0 mL container of helium gas at standard pressure is heated from 20.oC to 91oC and the pressure is increased to 2.8 atm, then what would the new volume be for the He gas

Answer 1

Answer: V₂ = 37.71mL

Explanation: To determine the new volume of Helium gas, use the Combined Gas Law, which states the following relationship among pressure, volume and temperature:

$\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}$

where index 1 relates to the initial state of the gas and index 2 to the final state of the gas.

Temperature is in Kelvin, so:

T = °C + 273

For this situation, standard pressure is 1 atm. Temperatures will be:

T₁ = 20 + 273 = 293 K

T₂ = 91 + 273 = 364 K

Solving:

$V_{2}=\frac{T_{2}P_{1}V_{1}}{P_{2}T_{1}}$

$V_{2}=\frac{364*1*85}{2.8*293}$

$V_{2}=\frac{30940}{820.4}$

$V_{2}=$ 37.71

The new volume of He gas is 37.71 mL.