Question

Air in a cylinder is compressed to one-tenth its original volume with no change in temperature. What is the
change in its pressure?

Answer 1

Answer on Question #49187 – Physics – Mechanics | Kinematics | Dynamics

1. Air in a cylinder is compressed to one-tenth its original volume with no change in temperature. What is the change in its pressure?

$$\frac{\eta_{1} = 0.1}{\eta_{2} - ?}$$

Solution.

Let write the ideal gas law for the initial and final state of the gas:

$$p_{1} V_{1} = \frac{m}{M} R T, \quad p_{2} V_{2} = \frac{m}{M} R T,$$

where $m$ and $M$ are the mass and molar mass, correspondingly. The gas temperature $T$ remains constant.

If we divide the first equation by the second one, the following equation will be obtained:

$$\frac{p_{1} V_{1}}{p_{2} V_{2}} = 1, \quad \text{or} \quad \frac{p_{2}}{p_{1}} = \frac{V_{1}}{V_{2}}.$$

According to the text, $\frac{V_{2}}{V_{1}} = 1 - \eta_{1}$. So, change in gas pressure is:

$$\eta_{2} = \frac{p_{2} - p_{1}}{p_{1}} = \frac{p_{2}}{p_{1}} - 1 = \frac{1}{\frac{V_{2}}{V_{1}}} - 1 = \frac{1}{1 - \eta_{1}} - 1 = \frac{\eta_{1}}{1 - \eta_{1}}.$$ $$\eta_{2} = \frac{0.1}{1 - 0.1} = \frac{1}{9}.$$

Answer: the pressure increases to $\frac{1}{9}$-th towards its original quantity (the pressure increases at $\frac{10}{9}$ times).