If #3/7 L# of a gas at room temperature exerts a pressure of #9 kPa# on its container, what pressure will the gas exert if the container's volume changes to #5/9 L#?

Question

Charlotte Aaron · Accepted Answer

#P_2 = 243/35 kPa# Let #P_1 = 9color(white).kPa# Let #V_1 = 3/7color(white).L# Let #V_2 = 9/5color(white).L# #P_1V_1 = P_2V_2# #P_2 = P_1V_1/V_2# #P_2 = 243/35 kPa#

Sebastian Acosta · Answer

#243/35#kPa

Boyle's Law for gases states that, given a constant temperature and mass, the variations in pressure and volume vary inversely. This can be expressed using the following formula: #P_1V_1=P_2V_2# #P_1=9#kPa #V_1=3/7 L# #P_2=#? #V_2=5/9 L# #9*3/7=P_2*5/9# #9/5*9*3/7=P_2*cancel(5)/cancel(9)*cancel(9)/cancel(5)color(white)(aaa)#Multiply both sides by 9/5 #243/35=P_2# #P_2=243/35#kPa

Benjamin Asiedu · Answer

undefined Use the combined gas law, which states:

$P_1V_1/T_1 = P_2V_2/T_2$, where $P_1$, $V_1$, and $T_1$ are the initial pressure, volume, and temperature, and $P_2$, $V_2$, and $T_2$ are the final pressure, volume, and temperature, respectively.

Given that the volume changes from $3/7$ L to $5/9$ L, and assuming the temperature remains constant (since room temperature is usually considered constant), we can solve for $P_2$.

$$
\begin{aligned}
P_1V_1 &= P_2V_2 \
(9 \text{ kPa})(3/7 \text{ L}) &= P_2(5/9 \text{ L}) \
\frac{27}{7} \text{ kPa} \cdot \text{L} &= \frac{5}{9} P_2 \text{ kPa} \cdot \text{L} \
\frac{27}{7} \text{ kPa} \cdot \text{L} \cdot \frac{9}{5} &= P_2 \
\frac{243}{35} \text{ kPa} &= P_2 \
\end{aligned}
$$

So, the gas will exert a pressure of $243/35$ kPa when the volume changes to $5/9$ L.

Charles Aeschliman · Answer

undefined To solve this problem, we can use Boyle's Law, which states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional. Mathematically, Boyle's Law is expressed as $P_1V_1 = P_2V_2$, where $P_1$ and $V_1$ are the initial pressure and volume, and $P_2$ and $V_2$ are the final pressure and volume, respectively.

Given that the initial volume is $3/7$ L and the initial pressure is 9 kPa, and the final volume is $5/9$ L, we can plug these values into Boyle's Law and solve for the final pressure ($P_2$):

$$
\frac{3}{7} \times 9 = P_2 \times \frac{5}{9}
$$

$$ \frac{27}{7} = \frac{5P_2}{9} $$

To find $ P_2 $, multiply both sides by $ \frac{9}{5} $:

$$ P_2 = \frac{27}{7} \times \frac{9}{5} $$

$$ P_2 = \frac{243}{35} $$

So, the pressure the gas will exert if the container's volume changes to $ \frac{5}{9} $ L is $ \frac{243}{35} $ kPa.