Question

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the same amount of air is placed in a box with a volume of 5.0m3 at 35∘C? Report your answer with two significant figures.

Answer 1

Answer:

Given:

• Initial pressure: $3.65\; \rm atm$.
• Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$.
• Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$.

New pressure: approximately $3.4\times 10\; \rm atm$ ($34\; \rm atm$.) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

• Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$. Calculate the new pressure, $P_1$.
• Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$. Calculate the final pressure, $P_2$.

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$.

Let $P_0$ denote the pressure of this gas before the volume change ($P_0 = 3.65\; \rm atm$.) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$:

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$.

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$.

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$.

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$.

Let $P_1$ denote the pressure of this gas before this temperature change ($P_1 = 32.85\; \rm atm$.) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$.

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$:

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$.

Calculate $P_2$, the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$.

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$.