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Problem 2. At a temperature of 300 K, the pressure of the gas in a deodorant can is 3 atm. Calculate
the pressure of the gas when it is heated to 900 K.​

Sagot :

Answer:

At a temperature of 300 K, the pressure of the gas in a deodorant can is 3 atm. Calculate the pressure of the gas when it is heated to 900 K. Therefore, final pressure (P2) = (P1T2)/T1 = (3 atm*900K)/300K = 9 atm.

Step-by-step explanation:

Given:

[tex]P_{1} = \text{3 atm}[/tex]

[tex]T_{1} = \text{300 K}[/tex]

[tex]T_{2} = \text{900 K}[/tex]

Required:

[tex]P_{2}[/tex]

Strategy:

This is a gas law problem. What gas law should we use?

Since the given quantities are pressure and temperature, we will use Gay-Lussac's law. According to this gas law, the pressure of a gas is directly proportional to its absolute temperature keeping the volume and the amount of gas constant.

Caution: The temperature must be converted to kelvin. If the given temperature is in degree Celsius, add 273 or 273.15 to convert it to kelvin.

The formula used for Gay-Lussac's law is

[tex]\boxed{\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}}[/tex]

where:

[tex]P_{1} = \text{initial pressure}[/tex]

[tex]T_{1} = \text{initial temperature}[/tex]

[tex]P_{2} = \text{final pressure}[/tex]

[tex]T_{2} = \text{final temperature}[/tex]

Solution:

Starting with the formula of Gay-Lussac's law

[tex]\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}[/tex]

Multiplying both sides of the equation by T₂ to solve for P₂

[tex]P_{2} = P_{1} \times \dfrac{T_{2}}{T_{1}}[/tex]

Substituting the given values and solving for P₂

[tex]P_{2} = \text{3 atm} \times \dfrac{\text{900 K}}{\text{300 K}}[/tex]

Therefore, the final pressure is

[tex]\boxed{P_{2} = \text{9 atm}}[/tex]

Answer:

P₂ = 9 atm

[tex]\\[/tex]

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