Question

A 100.0mL bubble of hot gases at 225 C and 1.80 atm escapes from an active volcano, what is the new volume of the bubble outside the volcano where the temperature is -25 C and the pressure is 0.80 atm

Answer 1

Answer:

112.08 mL

Explanation:

From the question we are given;

• Initial volume, V1 = 100.0 mL
• Initial temperature, T1 = 225°C, but K = °C + 273.15

thus, T1 = 498.15 K

• Initial pressure, P1 = 1.80 atm
• Final temperature , T2 = -25°C

                                     = 248.15 K

• Final pressure, P2 = 0.80 atm

We are required to calculate the new volume of the gases;

• According to the combined gas law equation;

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

Rearranging the formula;

$V2=\frac{P1V1T2}{T1P2}$

Therefore;

$V2=\frac{(1.80atm)(100mL)(248.15K)}{(498.15K)(0.80atm)}$

$V2=112.08mL$

Therefore, the new volume of the gas is 112.08 mL