The temperature of a sample of gas in steel container at 30.3 Atm is
2 answers:
Answer:
74.09 atm
Explanation:
Using the gas laws ( Charles and Boyle's law). We have the formula ,
P1/T1 = P2/T2
Where P1 = 30.3atm
T1 = -100 degree Celsius
to kelvin = -100+ 273 = 173K
T2 = 150 degree Celsius
To Kelvin = 150 = 150+273 = 423K
Imputing values
P1/T1 = P2/T2
30.3/173 = P2/ 423
Cross multiply
173×P2 = 30.3 ×423
173P2 = 12816.9
Divide both sides by 173
P2 = 12816.9/173
P2 = 74.09 atm
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Answer:
P2=74.086atm
Explanation:
Using the formula p1/t1=P2/t2
P1=30.3atm
P2=?
T1=-100°c to Kelvin=273-100=173k
T2=150°c to Kelvin=273+150=423k
30.3\173=x/423
Cross multiply
173x=423×30.3
173x=12816.9
Divide both sides by 173
X=74.086atm
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