What was the initial temperature of the air, if, when it was cooled by 10K, the volume decreased 2.5 times

What was the initial temperature of the air, if, when it was cooled by 10K, the volume decreased 2.5 times from the initial one, the process is considered isobaric.

Given:

T1 = T2 + 10 – the air was cooled by 10 degrees Kelvin;

V1 = 2.5 * V2 – the air volume has decreased 2.5 times;

P1 = P2 – isobaric process.

It is required to determine the initial air temperature T1 (Kelvin).

Since, according to the condition of the problem, the process was isobaric, then:

V1 / T1 = V2 / T2;

V1 / T1 = V1 / (2.5 * (T1 – 10));

V1 / T1 = V1 / (2.5 * T1 – 25);

2.5 * T1 – 25 = T1;

2.5 * T1 – T1 = 25;

1.5 * T1 = 25;

T1 = 25 / 1.5 = 16.6 degrees Kelvin.

Answer: the initial air temperature was 16.6 degrees Kelvin.