The gas inside of a container exerts #9 Pa# of pressure and is at a temperature of #250 ^o K#. If the pressure in the container changes to #5 Pa# with no change in the container's volume, what is the new temperature of the gas?

Question

The gas inside of a container exerts #9   Pa# of pressure and is at a temperature of #250 ^o K#.  If the pressure in the container changes to #5 Pa# with no change in the container's volume, what is the new temperature of the gas?

Leo Abeyta · Accepted Answer

Approximately #138.9# kelvin.

For constant volume and number of moles, we can use Gay-Lussac's law, which states that, #PpropT# or #P_1/P_2=T_1/T_2#. So here, we get: #(9 \ "Pa")/(250 \ "K")=(5 \ "Pa")/T_2# #T_2=(250 \ "K")/(9color(red)cancelcolor(black)"Pa")*5color(red)cancelcolor(black)"Pa"# #~~138.9 \ "K"#

Nicholas Agrusa · Answer

undefined Using the ideal gas law, $PV = nRT$, where:
- $P$ is the pressure (in pascals),
- $V$ is the volume (in cubic meters),
- $n$ is the number of moles of gas,
- $R$ is the ideal gas constant ($8.31 \, \text{J/mol} \cdot \text{K}$),
- $T$ is the temperature (in kelvin).

Since the volume remains constant, we can simplify the equation to $P_1 / T_1 = P_2 / T_2$ where $P_1$, $T_1$ are the initial pressure and temperature, and $P_2$, $T_2$ are the final pressure and temperature.

Rearranging the equation to solve for $T_2$, we get:
$$T_2 = \frac{P_2 \cdot T_1}{P_1}$$

Substituting the given values, we get:
$$T_2 = \frac{5 \, \text{Pa} \times 250 \, \text{K}}{9 \, \text{Pa}}$$

After calculating, we find the new temperature of the gas.

Lily Adams · Answer

undefined The new temperature of the gas is 139.58 K.