The gas inside of a container exerts #9 Pa# of pressure and is at a temperature of #690 ^o K#. If the pressure in the container changes to #34 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 #690 ^o K#.  If the pressure in the container changes to #34 Pa# with no change in the container's volume, what is the new temperature of the gas?

Leo Abeyta · Accepted Answer

The temperature is #=2606.7K# We use Gay Lussac's Law #P_1/T_1=P_2/T_2# #P_1=9Pa# #T_1=690K# #P_2=34Pa# Therefore, #T_2=(P_2T_1)/P_1# #=(34*690)/9=2606.7K#

Carson Alexander · Answer

undefined To find the new temperature of the gas when the pressure changes, you can use the ideal gas law equation:

$$ P_1 \cdot V = P_2 \cdot V $$

where:
- $ P_1 $ is the initial pressure (9 Pa),
- $ P_2 $ is the final pressure (34 Pa),
- $ V $ is the volume (constant).

Since the volume is constant, it cancels out from the equation. Solving for the new temperature $ T_2 $:

$$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$

$$ T_2 = \frac{P_2 \cdot T_1}{P_1} $$

Substituting the given values:

$$ T_2 = \frac{34 \cdot 690}{9} $$

$$ T_2 \approx 2620 \, \text{K} $$

Therefore, the new temperature of the gas is approximately 2620 Kelvin.