Question

A gas was initially held at -50.00 °C at a pressure of 25 atm. After heating the gas to 30.00 °C, the pressure decreased to 15 atm, and the volume of gas was found to be 25 L. What was the initial volume of the gas? • Use --273.15 °C for absolute zero. • Report your answer with two significant figures.

Answer 1

By gas equation  

$\frac{P_1V_1}{T_1}$ = $\frac{P_2V_2}{T_2}$ ... (1)

Given, initial pressure( P₁ ) = 25 atm, initial temperature ( T₁ )

= (- 50 + 273.15) K = 223.15 K.

let initial volume is V₁

now, final volume (V₂) = 25 L.

Final pressure (P₂) = 15 atm

Final temperature (T₂) = ( 30.00 + 273.15 ) K = 303.15 K.

using Eq.1

  
25 x V1 223.15
=
15 x 25 303.15

Or, V₁ = (15×25×223.15)/(25×303.15) L.

Or, V₁ = 11 L (2 significant figures)

So, initial volume of the gas was 11 L.