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Cloud [144]
3 years ago
15

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
Chemistry
1 answer:
Inessa05 [86]3 years ago
8 0
<h3>Answer:</h3>

112.08 mL

<h3>Explanation:</h3>

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

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