The temperature of 2.5 L of a gas initially at STP is raised to 250 °C at consta...

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The temperature of 2.5 L of a gas initially at STP is raised to 250 °C at constant volume. Calculate the final pressure of the gas in atm.

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Parthenogenesis

✏️GAY-LUSSAC'S LAW

$\sf{\large{\red{STRATEGY:}}}$

We are given the initial temperature, the initial pressure, and the final temperature; we are asked to solve for the final pressure of a gas. Since we assume that the volume and the amount of a gas are constant, we use Gay-Lussac's law.

Gay-Lussac's law is given by

$\bold{\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}}$

where:

P₁ = initial pressure

T₁ = initial temperature (in kelvin)

P₂ = final pressure

T₂ = final temperature (in kelvin)

Note 1: If the given temperature is in °C, we must convert it to kelvin (K) first before substituting in the formula.

Note 2: At STP, the temperature and pressure of an ideal gas are 273 K and 1.0 atm, respectively

$\sf{\large{\red{SOLUTION:}}}$

Listing the given values, we have

$\begin{aligned} P_1 & = \text{1.0 atm} \\ T_1 & = \text{273 K} \\ T_2 & = 250 + 273 = \text{523 K} \end{aligned}$

Arranging the equation, we can now solve for the final pressure of the gas. Therefore,

$\begin{aligned} P_2 & = \dfrac{P_1T_2}{T_1} \\ & = \dfrac{(\text{1.0 atm})(\text{523 K})}{\text{273 K}} \\ & = \boxed{\text{1.9 atm}}\end{aligned}$

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