Question

The temperature of a sample of gas in steel container at 30.3 Atm is

Answer 1

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 2

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