Answer:
[tex]\boxed {\boxed {\sf P_2=2.03 \ atm}}[/tex]
Explanation:
We are concerned with the variables of temperature and pressure, so we use Gay-Lussac's Law, which states the temperature of a gas is directly proportional to the pressure. The formula is:
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
We know that the container of gas begins at standard temperature and pressure (STP). This is 1 atmosphere of pressure and 273 Kelvin.
[tex]\frac { 1 \ atm}{ 273 \ K} = \frac{P_2}{T_2}[/tex]
We know the gas is put into an oven at 280 degrees Celsius. We can convert this to Kelvin.
- K= °C + 273.15
- K= 280 +273.15
- K= 553.15
[tex]\frac { 1 \ atm}{ 273 \ K} = \frac{P_2}{553.15 \ K}[/tex]
We are solving for the new pressure, so we must isolate the variable P₂. It is being divided by 553.15 Kelvin. The inverse of division is multiplication, so we multiply both sides by 553.15 K
[tex]553.15 \ K *\frac { 1 \ atm}{ 273 \ K} = \frac{P_2}{553.15 \ K} * 553.15 \ K[/tex]
[tex]553.15 \ K *\frac { 1 \ atm}{ 273 \ K}= P_2[/tex]
The units of Kelvin cancel.
[tex]553.15 *\frac { 1 \ atm}{ 273 }= P_2[/tex]
[tex]2.02619047619 \ atm = P_2[/tex]
Rounded to the nearest hundredth:
[tex]2.03 \ atm \approx P_2[/tex]
The new pressure is approximately 2.03 atmospheres.
