A lungful of air (377 cm3 ) is exhaled into a machine that measures lung capacity. If the air is exhaled from the lungs at a pressure of 1.7 atm at 41.7 ◦C but the machine is at ambient conditions of 0.958 atm and 23 ◦C, what is the volume of air measured by the machine? Answer in units of cm3 .

Respuesta :

Answer: The volume of the gas when the pressure and temperature has changed is [tex]629.2cm^3[/tex]

Explanation:

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

The equation follows:

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas

[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas

We are given:

[tex]P_1=1.7atm\\V_1=377cm^3\\T_1=41.7^oC=[41.7+273]K=314.7K\\P_2=0.958atm\\V_2=?cm^3\\T_2=23^oC=[23+273]K=296K[/tex]

Putting values in above equation, we get:

[tex]\frac{1.7atm\times 377cm^3}{314.7K}=\frac{0.958atm\times V_2}{296K}\\\\V_2=\frac{1.7\times 377\times 296}{314.7\times 0.958}=629.2cm^3[/tex]

Hence, the volume of the gas when the pressure and temperature has changed is [tex]629.2cm^3[/tex]