Question

What is the final temperature, in degrees Celsius, of the gas in the bubble outside the volcano if the final volume of the bubble is 130. mL and the pressure is 0.760 atm , if the amount of gas does not change?

Answer 1

The question is incomplete, here is the complete question:

A 127 mL bubble of hot gases at 213°C and 1.89 atm is emitted from an active volcano.

What is the final temperature, in degrees Celsius, of the gas in the bubble outside the volcano if the final volume of the bubble is 136 mL and the pressure is 0.850 atm , if the amount of gas does not change?

Answer: The temperature of the gas is -40°C

Explanation:

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=1.89atm\\V_1=127mL\\T_1=213^oC=[213+273]K=484K\\P_2=0.850atm\\V_2=136mL\\T_2=?$

Putting values in above equation, we get:

$\frac{1.89atm\times 127mL}{484K}=\frac{0.850atm\times 136mL}{T_2}\\\\T_2=\frac{0.850\times 136\times 484}{1.89\times 127}=233K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$233=T(^oC)+273\\\\T(^oC)=-40^oC$

Hence, the temperature of the gas is -40°C