Question

The temperature of a sample of gas in a steel container at 30.0 kpa is increased from -100.0 °c to 1.00x10 the the power of 3 °c what is the final pressure inside the tank ?

Answer 1

Step 1 - Understanding how pressure and temperature are related

There are three main variable that may modify the state of a gas: pressure (P), volume (V) and temperature (T). They are all related by the following formula:

`pV=nRT`

When one of these variables is kept constant (in this exercise, for example, volume is constant), linear relationships arise between the remaining variables.

So, at constant volume, presure and temperature become directly proportional to each other, which can be mathematically stated as:

`(p_1)/(T_1)=(p_2)/(T_2)`

We can use this relation to solve the exercise.

Step 2 - Solving the exercise

According to the exercise:

`\begin{gathered} p_1=30\text{ kPa} \\ \\ T_1=-100°C\text{ \lparen173K\rparen} \\ \\ p_2=? \\ \\ T_2=1000°C(1273K) \end{gathered}`

For this relation to work, the temperature must be used in K, not °C.

Setting the values in the equation, we get:

`\begin{gathered} (30)/(173)=(p__2)/(1273) \\ \\ p_2=(1273*30)/(173)=220.7\text{ kPa} \end{gathered}`

Answer: the final pressure in the tank is 220.7 kPa.