Acylinder of gas has a pressure of 440 kpa at 25*c. at what temperature in *c will it reach 650 kpa

              The answer is 167.29 °C

Explanation:

According to Gay-Lussac's Law: The temperature and Pressure of a given gas are directly proportional to each other at given volume and amount.

The formula is as:

                                                 P₁ / T₁  =  P₂ / T₂   (1)

In given case:

                       T₁  =  25 °C  =  298.15 K

                       T₂  =  Unknown

                       P₁  =  440 kPa

                       P₂  =  650 kPa

Solving equation 1 for T₂,

                                       T₂  =   P₂ × T₁ / P₁

Putting Values,

                                       T₂  =   650 kPa × 298.15 K / 440 kPa

                                       T₂  =   440.448 K

Or,

                                      T₂  =   167.29 °C                 ∴ °C = K - 273.15