Question

A gas originally at a temperature of $26.5°C$ is cooled at constant pressure. Its volume decreases from 5.40 L to 5.26 L. Determine its new temperature in degrees Celsius.

Short answer

Thus, the new temperature in degrees Celsius is $18.73°C$

Step-by-step solution

Charles Law:

The Charles law is one of the expressions of the combined gas law. According to this law at constant pressure, the volume of the gas is directly proportional to the temperature of the gas.

Given Data

Temperature = $26.5°C$

Volume decreases from 5.40 L to 5.26 L.

Calculation

At constant pressure, the formula is

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Here,

* V₁ is the initial volume of the gas, i.e., 5.40 L.

* V₂ is the final volume of the gas, i.e., 5.26 L.

* T₁ is the initial temperature of the gas, i.e.,$26.5°C(26.5+273.15)K=299.65K$

* T₂ is the final temperature of the gas.

Rearranging and substituting these values, in the given formula we, get

$$\begin{gathered} T_2=\frac{V_2\times T_1}{V_1} \\ =\frac{5.26L\times 299.65K}{5.40 L} \\ =291.88 K \end{gathered}$$

On converting the K temperature to degrees Celsius, we need to subtract 273.15 from the given temperature as:

T₂₌(291.88K-273.15)$°$C

=18.73$°$C

Thus, at 18.73$°$C temperature the volume will decrease to 5.26 L