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Chapter 5: Problem 36

The temperature of \(2.5 \mathrm{~L}\) of a gas initially at STP is increased to \(250^{\circ} \mathrm{C}\) at constant volume. Calculate the final pressure of the gas in atm.

Short Answer

Expert verified
Computing the equation above, the final pressure \(\(P2\)\) turns out to be approximately \(1.915 atm\). So, the final pressure of the gas after being heated to \(250^{\circ} \mathrm{C}\) at constant volume is approximately \(1.915 atm\).

Step by step solution

01

Understand the problem and data

You initially have a gas at STP, meaning the gas is at a standard temperature of \(0^{\circ} C\) (or \(273.15 K\)) and a standard pressure of \(1 atm\). The temperature is then raised to \(250^{\circ} C\), while the volume remains constant, and the final pressure is to be found.
02

Convert temperatures to the Kelvin scale

The formulae in gas laws typically involve absolute temperature, meaning the Kelvin scale must be used. To convert from Celsius to Kelvin, we add \(273.15\) to the Celsius temperature. That's why the initial temperature \(0^{\circ} C\) becomes \(0 + 273.15 = 273.15 K\) and the final temperature \(250^{\circ} C\) becomes \(250 + 273.15 = 523.15 K\).
03

Apply Gay-Lussac's Law

According to Gay-Lussac's Law, the pressure of a gas is directly proportional to its temperature if its volume is constant. Hence, \(\frac{P1}{T1} = \frac{P2}{T2}\). Given that \(P1 = 1 atm\), \(T1 = 273.15 K\), \(T2 = 523.15 K\), we need to solve for \(P2 = \frac{P1 × T2}{T1}\).
04

Calculate final pressure

Substituting the known values into the formula, you get \(P2 = \frac{1 atm × 523.15 K}{273.15 K}\). Solving this calculation will provide the final pressure value.

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