At a temperature of 273 K, a 400.-milliliter gas sample has a pressure of 760. millimeters of mercury. If the pressure is changed to 380. millimeters of mercury, at which temperature will this gas sample have a volume of 551 milliliters

188.03K

Explanation:

The following data were obtained from the question:

T1 (initial temperature) =273K

V1 (initial volume) = 400mL

P1 (initial pressure) = 760mmHg

P2 (final pressure) = 380mmHg

V2 (final volume) = 551mL

T2 (final temperature) =?

Applying the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas can be obtained as follow:

P1V1/T1 = P2V2/T2

760 x 400/273 = 380 x 551/T2

Cross multiply to express in linear form as shown below:

760 x 400 x T2 = 273 x 380 x 551

Divide both side by 760 x 400

T2 = (273 x 380 x 551)/(760 x 400)

T2 = 188.03K.

Therefore, at a temperature of 188.03K, the gas sample will have a volume of 551mL

what is the question?

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.