Answer:
[tex]T_{2}=278.80 K [/tex]
Explanation:
Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.
[tex] (\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}[/tex].
Now, let's use the ideal gas equation to the initial and the final state:
[tex]\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}[/tex]
Let's recall that the term nR is a constant. That is why we can match these equations.
We can find a relation between the volumes of the initial and the final state.
[tex] \frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}[/tex]
Combining this equation with the first equation we have:
[tex](\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}[/tex]
[tex](\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}[/tex]
Now, we just need to solve this equation for T₂.
[tex]T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2} [/tex]
Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.
Here,
[tex] p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40 [/tex]
Finally, T2 will be:
[tex]T_{2}=278.80 K [/tex]
