The gas inside of a container exerts #25 Pa# of pressure and is at a temperature of #450 ^o K#. If the temperature of the gas changes to #150 ^oC# with no change in the container's volume, what is the new pressure of the gas?

Question

The gas inside of a container exerts #25   Pa# of pressure and is at a temperature of #450 ^o K#.  If the temperature of the gas changes to #150 ^oC# with no change in the container's volume, what is the new pressure of the gas?

Charlotte Aaron · Accepted Answer

The pressure is #=23.5Pa# We use Gay Lussac's Law #P_1/T_1=P_2/T_2# #P_1=25 Pa# #T_1=450K# #T_2=150+273=423K# #P_2=P_1/T_1*T_2=25/450*423=23.5Pa#

Xavier Adame · Answer

undefined To find the new pressure of the gas, we can use the ideal gas law, $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles of gas, $R$ is the ideal gas constant, and $T$ is temperature in Kelvin.

Since the volume of the container remains constant, we can simplify the equation to $\frac{P_1}{T_1} = \frac{P_2}{T_2}$, where $P_1$ and $T_1$ are the initial pressure and temperature, and $P_2$ and $T_2$ are the final pressure and temperature, respectively.

Given:
$P_1 = 25 \, \text{Pa}$
$T_1 = 450 \, \text{K}$
$T_2 = 150 \, ^\circ C = 150 + 273 = 423 \, \text{K}$

Now, we can rearrange the equation to solve for $P_2$:
$P_2 = \frac{P_1 \times T_2}{T_1}$

Substituting the given values:
$P_2 = \frac{25 \times 423}{450}$

$P_2 ≈ 23.33 \, \text{Pa}$

So, the new pressure of the gas is approximately $23.33 \, \text{Pa}$.