If a gas at 35.0 °c is pressurized from 160 kpa to 250 kpa, determine the final temperature of the gas. assume the volume and amount of gas are held constant. show your work.

This problem is a simple application of Gay-Lussac's Law for gases.

Gay - Lussac's Law states that the pressure and temperature (in absolute scale) of a gas are proportional.

This is:

p / T = constant => p1 / T1 = p2 / T2.

Here you have p1 = 160 kPa; T1 = 35.0 + 273.15 = 308.15 K; p2 = 250 KPa; T2 = ?

Then, just solve T2 from the equation above.

T2 = p2 * T1 / p1 =   250 kPa * 308.15K / 160 kPa = 481.48 K.

T2 = 481.48 - 273.15 = 208.33 °C.

208.33 °C.