Question

An empty pressure cooker of volume $$10$$ litres contains air at atmospheric pressure $${10}^5\text{Pa}$$ and temperature of $$27{}^{\circ }\text{C}$$ . It contains a whistle which has area of $$0.1\text{c}{\text{m}}^2$$ and weight of $$100\text{g}$$ . What should be the temperature of air inside so that the whistle is just lifted up?

Answer 1

Hint:First of all, we will find the pressure by using the formula of force. After that we will use the relation between the temperature and pressure with constant volume. We will manipulate accordingly and find the result.

Complete step by step answer:
In the given question, we are supplied with the following data:
The net volume of the pressure cooker is $$10$$ litres. The pressure of the air inside the pressure cooker is atmospheric air $${10}^5\text{Pa}$$ .The temperature inside the cooker is $$27{}^{\circ }\text{C}$$ .The whistle which is present at the top of the cooker has an area of $$0.1\text{c}{\text{m}}^2$$ and weight of $$100\text{g}$$ .We are asked to find the minimum temperature of air inside so that the whistle is just lifted.

To begin with, we know that the whistle of the pressure cooker can be lifted only by the pressure of the air inside. The whistle has got mass and the force needs to act along the upward direction in order to lift the whistle. Let us convert the mass and area in the S.I system of units.
Mass of the whistle:
$$\begin{gathered} 100\text{g} \\ \Rightarrow 100\times {10}^{-3}\text{kg} \\ \Rightarrow 0.1\text{kg} \end{gathered}$$
Area of the whistle:
$$\begin{gathered} 0.1\text{c}{\text{m}}^2 \\ \Rightarrow 0.1\times {\left({10}^{-2}\right)}^2{\text{m}}^2 \\ \Rightarrow {10}^{-5}{\text{m}}^2 \end{gathered}$$
Let the pressure that is required to lift the whistle be $$P$$ .
We can write as:
$$m\times g=P\times A$$ …… (1)
Where,
$$m$$ indicates the mass of the whistle.
$$A$$ indicates the area of the whistle.

Now we substitute the required values in the equation (1) and we get:
$$\begin{gathered} m\times g=P\times A \\ \Rightarrow 0.1\times 9.8=P\times {10}^{-5} \\ \Rightarrow P=0.98\times {10}^5\text{Pa} \end{gathered}$$
Since, the volume is constant we can use the relation between temperature and pressure, which is given as:
$$\begin{gathered} {\displaystyle \frac{P_0}{T_1}}={\displaystyle \frac{P}{T}} \\ \Rightarrow {\displaystyle \frac{{10}^5}{27{}^{\circ }\text{C}}}={\displaystyle \frac{0.98\times {10}^5}{T}} \\ \Rightarrow {\displaystyle \frac{{10}^5}{273+27}}={\displaystyle \frac{0.98\times {10}^5}{T}} \\ \Rightarrow T=294\text{K} \end{gathered}$$
Hence, the minimum temperature of air inside so that the whistle is just lifted is $$294\text{K}$$ .
We can convert it into degree centigrade:
$$\begin{gathered} 294\text{K} \\ \Rightarrow \left(294-273\right){}^{\circ }\text{C} \\ \therefore 21{}^{\circ }\text{C} \end{gathered}$$
Hence, the temperature of air inside so that the whistle is just lifted up is $21{}^{\circ }\text{C}21{}^{\circ }\text{C}$.

Note: While solving the problem, always remember that we need to convert the given temperature into kelvin scale. On failure to do so, the answer will come truly irrelevant and wrong. If you want you can use the temperature in degree centigrade in the above formula and see the difference.