Question

After conditions changed to a volume of a sample of helium at 15.56 mL, 138.7°C and 334.6 kPa. What was its initial volyme at 63.2 °C and 57.3 kPa?O a. 74.2O b. 41.4O c. 2.18O d. 111

Answer 1

a. 74.2

Step-by-step explanation

Given that:

The initial temperature, T₁ = 63.2 °C + 273 = 336.2 K

Initial pressure, P₁ = 57.3 kPa

The final volume, V₂ = 15.56 mL

Final temperature, T₂ = 138.7°C + 273 = 411.7 K

Final pressure, P₂ = 334.6 kPa

What to find:

The initial volume, V₁.

Step-by-step solution:

The initial volume, V₁ can be calculated using the combined gas law equation.

`\begin{gathered} (P_1V_1)/(T_1)=(P_2V_2)/(T_2) \\ \\ V_1=(P_2V_2T_1)/(P_1T_2)=(334.6kPa*15.56mL*336.2K)/(57.3kPa*411.7K) \\ \\ V_1=\frac{1750383.611\text{ }mL}{23590.41}=74.2\text{ }mL \end{gathered}`

Hence, its initial volume at 63.2 °C and 57.3 kPa is 74.2 mL