A gas occupies a volume of 0.444 L at 0.00C and 79.00 kPa. What is the final Kelvin temperature when
the volume of the gas is changed to 1880. mL and the
pressure is changed to 38.70 kPa?
Include unit of measurement and use proper
significant figures.

The final temperature is 566°K

Explanation:

Given

V1 = 0.444L

T1 = 0.00°C

P1 = 79.00 kPa

V2 = 1880mL

P2 = 38.70 kPa

Required

Determine the final temperature of the gas

To answer this, we make use of idea gas law equation.

This is:

(P1V1)/T1 = (P2V2)/T2

Convert V1 to mL

V1 = 0.444L

V1 = 0.444 * 1000mL

V1 = 444mL

Convert T1 to Kelvin

T1 = 0.00°C

T1 = 0.00 + 273K

T1 = 273K

Substitute these values in the given equation.

(P1V1)/T1 = (P2V2)/T2

(79 * 444)/273 = (38.70 * 1880)/T2

35076/273 = 72756/T2

Cross Multiply

35076 * T2 = 273 * 72756

35076T2 = 19862388

Make T2 the subject

T2 = 19862388 ÷ 35076

T2 = 566K (approximated)

Hence, the final temperature is 566°K