A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0°C. During the summer, the temperature

A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0°C. During the summer, the temperature in the storage area reached 40.0°C. What was the pressure in the gas tank then?

Answer:
Since the volume is kept constant we use Charles' law
p1/T1 = p2 / T2

T1 = 5 + 273 = 278 K
T2 = 40 + 273 =313 K

5.00 / 278 = p2 / 313
p2 = 5.63 atm
P1 x T2 = P2 x T1 (Definitely...Gay Lussac's Law).
5atm x 313K = P2 x 278K
P2 = (5 x 313) ÷ 278
P2 (New Pressure) = 5.63atm.

As P & T are directly proportional at constant volume...then...
Another method is: -
313K ÷ 278K = 1.126 times the initial temp.
5atm. x 1.126 = 5.63atm.

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