Respuesta :
Answer: The inside volume of the gas when the pressure and temperature has changed is 232.7 L
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=1.00atm\\V_1=4.25\times 10^4L\\T_1=15^oC=[15+273]K=288K\\P_2=175atm\\V_2=?\\T_2=3^oC=[3+273]K=276K[/tex]
Putting values in above equation, we get:
[tex]\frac{1atm\times 4.25\times 10^4L}{288K}=\frac{175atm\times V_2}{276K}\\\\V_2=\frac{1\times 4.25\times 10^4\times 276}{288\times 175}=232.7L[/tex]
Hence, the inside volume of the gas when the pressure and temperature has changed is 232.7 L
The volume of the gas inside when the container breaks under the pressure at this depth is 232.7L
According to the general gas equation is expressed as:
[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
[tex]P_1 \ and \ P_2[/tex] are the pressure of the gas:
[tex]V_1 \ and \ V_2[/tex] are the volume of the gas
[tex]T_1 \ and \ T_2[/tex] are the temperatures
Given the following parameters
P₁ = 1.00atm
P₂ = 175atm
V₁ = [tex]4.25 \times 10^4L[/tex]
V₂ = ?
T₁ = 15 + 273 = 288K
T₂ = 3 + 273 = 276K
Substitute the given parameters into the formula:
[tex]V_2=\frac{P_1V_1T_2}{P_2T_1} \\V_2=\frac{1\times42500\times276}{175\times288}\\V_2=232.7L[/tex]
Hence the volume of the gas inside when the container breaks under the pressure at this depth is 232.7L
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