The gas inside of a container exerts #27 Pa# of pressure and is at a temperature of #180 ^o K#. If the pressure in the container changes to #40 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 #27   Pa# of pressure and is at a temperature of #180 ^o K#.  If the pressure in the container changes to #40 Pa# with no change in the container's volume, what is the new temperature of the gas?

Layla Ahmed · Accepted Answer

Approximately #267# kelvin.

We can apply Gay-Lussac's law, which states that at a constant number of moles and volume, to #PpropT# or #P_1/T_1=P_2/T_2# Thus, by entering our supplied values into the formula, we can obtain, #(27 \ "Pa")/(180 \ "K")=(40 \ "Pa")/T_2# #=>T_2=(180 \ "K")/(27color(red)cancelcolor(black)"Pa")*40color(red)cancelcolor(black)"Pa"# #=266.bar(6) \ "K"# #~~267 \ "K"#

Axel Acevedo · Answer

undefined Use the ideal gas law: $P_1/T_1 = P_2/T_2$

$T_2 = (P_2/P_1) \times T_1$

$T_2 = (40/27) \times 180$

$T_2 ≈ 266.67 \, K$

Nicholas Agrusa · Answer

undefined Using the ideal gas law $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature in Kelvin, we can solve for the new temperature ($T_2$):

$$P_1V = nRT_1$$
$$P_2V = nRT_2$$

Since the volume ($V$) and the number of moles ($n$) remain constant:

$$P_1T_1 = P_2T_2$$

Substituting the given values:

$$27 \times 180 = 40 \times T_2$$

$$T_2 = \frac{27 \times 180}{40}$$

$$T_2 = \frac{4860}{40}$$

$$T_2 = 121.5 \, K$$

So, the new temperature of the gas is $121.5 \, K$.