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s344n2d4d5 [400]
3 years ago
14

a steel cylinder is left out in the sun the pressure at 2:00 pm was 1014 kpa at 33.0 c at 7:00 am at a temperature of 21.4c what

was the pressure brainly
Chemistry
1 answer:
e-lub [12.9K]3 years ago
5 0
The easiest way is to use the Law of Gay-Lussac. This law states that there is a direct relation between the temperature in Kelvin of a gas and the pressure.

Then, namig p the pressure and T the temperature in Kelvin and using subscripts for every state:

p/T is constant ==> p_1 / T_1 = p_2/T_2

From which you obtain:

p_2 = [p_1 / T_1] * T_2

T_1 = 33.0 + 273.15 = 306.15 K
T _2 = 21.4 + 273.15 = 294.55 K

p_1 = 1014 kPa

p_2 = 1014 kPa * 294.55 K / 306.15 K = 975.6 kPa
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