Question

Combined gas law problem: A balloon is filled with 500.0 mL of helium at a temperature of 27°Cand 755 mm Hg. As the balloon rises in the atmosphere, the pressure and temperature drop. What\·olume will it have when it reaches an altitude where the temperature is -33°C and the·pressure is0.65 atm'

Answer 1

The volume will be 0.61 L or 610 mL

Step-by-step explanation

Given:

Initial volume, V₁ = 500.0 mL = 0.5 L

Initial temperature, T₁ = 27 °C = (27 + 273.15) = 300.15 K

Initial pressure, P₁ = 755 mmHg = 0.993 atm

Final temperature, T₂ = -33 °C = 240.13 K

Final pressure, P₂ = 0.65 atm

What to find:

The final volume,, V.

Solution:

The combined gas law equation is given by:

`\begin{gathered} (P_1V_1)/(T_1)=(P_2V_2)/(T_2) \\ \\ \end{gathered}`

Put the values of the parameters into the combined law equation:

`\begin{gathered} (0.993atm*0.5L)/(300.15K)=\frac{0.65atm* V_2}{240.15\text{ }K} \\ \\ \frac{0.4965\text{ }atm.L}{300.15\text{ }K}=\frac{0.65atm* V_2}{240.15\text{K}} \\ \\ Cross\text{ }multiply \\ \\ V_2*0.65atm*300.15\text{ }K=240.15K*0.4965\text{ }atm.L \\ \\ V_2=\frac{240.15K*0.4965\text{ }atm.L}{0.65atm*300.15K} \\ \\ V_2=0.611\text{ }L \end{gathered}`

The volume will be 0.61 L or 610 mL