A gas is brought to a final pressure of 6.8 atm. after increasing its temperature from 40 K to 280 K. Calculate the original pressure of the gas.

0.97 atm.

Explanation:

From the question given above, the following data were obtained:

Final pressure (P2) = 6.8 atm

Initial temperature (T1) = 40 K

Final temperature (T2) = 280 K

Initial pressure (P1) =?

Thus, we can obtain the initial (original) pressure of the gas as follow:

P1/T1 = P2/T2

P1 /40 = 6.8/280

Cross multiply

P1 × 280 = 40 × 6.8

P1 × 280 = 272

Divide both side by 280

P1 = 272/280

P1 = 0.97 atm

Therefore, the original pressure of the gas is 0.97 atm.

first one is the correct answer

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