Certainly! To solve this problem step-by-step, we need to use the relation given by the gas law [tex]\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)[/tex], where:
- [tex]\(P_1\)[/tex] is the initial pressure.
- [tex]\(T_1\)[/tex] is the initial temperature (in Kelvin).
- [tex]\(P_2\)[/tex] is the final pressure.
- [tex]\(T_2\)[/tex] is the final temperature (in Kelvin).
Let's start with converting the given temperatures to Kelvin:
1. Convert the initial temperature [tex]\(T_1\)[/tex] from Celsius to Kelvin:
Given [tex]\(T_1 = 25^\circ C\)[/tex],
[tex]\[
T_1 = 25 + 273.15 = 298.15 \text{ K}
\][/tex]
2. Use the gas law to solve for the final temperature [tex]\(T_2\)[/tex] in Kelvin:
We are given the pressures and the relation:
[tex]\[
\frac{P_1}{T_1} = \frac{P_2}{T_2}
\][/tex]
Rearranging this equation to solve for [tex]\(T_2\)[/tex]:
[tex]\[
T_2 = \frac{P_2 \cdot T_1}{P_1}
\][/tex]
3. Substitute the values into the equation:
Given [tex]\(P_1 = 0.96 \text{ atm}\)[/tex], [tex]\(P_2 = 1.25 \text{ atm}\)[/tex], and [tex]\(T_1 = 298.15 \text{ K}\)[/tex],
[tex]\[
T_2 = \frac{1.25 \cdot 298.15}{0.96}
\][/tex]
[tex]\[
T_2 = \frac{372.6875}{0.96} \approx 388.21614583333337 \text{ K}
\][/tex]
4. Convert the final temperature [tex]\(T_2\)[/tex] from Kelvin back to Celsius:
[tex]\[
T_2 \text{ in Celsius} = T_2 - 273.15 = 388.21614583333337 - 273.15 = 115.0661458333334^\circ C
\][/tex]
So, the new temperature of the gas is approximately [tex]\(388.216 \text{ K}\)[/tex] or, when converted to Celsius, likewise approximately [tex]\(115.066 \text{ °C}\)[/tex].
Thus, from the given choices, the closest correct answer is:
[tex]\[
115^\circ C
\][/tex]
