At a temperature of 47C, the pressure in the vessel was 80 kPa. What will the pressure become

At a temperature of 47C, the pressure in the vessel was 80 kPa. What will the pressure become when the temperature drops by 9%

Given:

T1 = 47 degrees Celsius = 320 degrees Kelvin – the initial temperature of the gas in the vessel;

P1 = 80 kPa = 80,000 Pascal – initial gas pressure in the vessel;

T2 = T1 – 9% – gas pressure decreased by 9%.

It is required to determine P2 (Pascal) – the final gas pressure in the vessel.

According to the condition of the problem, the process takes place at a constant volume.

Then, we find the final gas temperature:

T2 = T1 – 9% = 320 – 0.09 * 320 = 320 – 28.8 = 291.2 degrees Kelvin.

P1 / T1 = P2 / T2, from here we find:

P2 = P1 * T2 / T1 = 80,000 * 291.2 / 320 = 80,000 * 0.91 = 72800 Pascal = 72.8 kPa.

Answer: the final gas pressure is 72.8 kPa.