Gas at 27 degrees occupies a volume of 0.004m3. with isobaric heating of a given mass of gas

Gas at 27 degrees occupies a volume of 0.004m3. with isobaric heating of a given mass of gas, its volume will increase to 0.017m3. Find the end temperature?

Given:

t1 = 27 oC

V1 = 0.004 m3

V2 = 0.017 m3

p = const

m = const

Find: t2 =?

Decision:

Let’s convert the gas temperature to the absolute temperature on the Kelvin scale.

Absolute gas temperature T = t + 273 K.

Then T1 = 27 oC + 273K = 300 K.

In the case of an isobaric process according to the Gay-Lussac law

for m = const at p = const V1 / T1 = V2 / T2.

Hence, T2 = V2 * T1 / V1, T2 = 0.017 m3 * 300 K / 0.004 m3 = 1275 K.

On the Celsius scale, the temperature is t2 = T2 – 273 K = 1275 K – 273 K = 1002 oC.

Answer: As a result of isobaric heating, the final gas temperature will be 1002 oC.