Helium gas in a 2.00 L cylinder is under 1.20 atm pressure. At 40.5 degree Celsius that same gas sample has a pressure of 2.00 atm. What was the initial temperature of the gas in the cylinder? (Amonton's Law)

Sagot :

Given:

[tex]P_{2} = \text{2.00 atm}[/tex]

[tex]T_{2} = \text{40.5°C + 273.15 = 313.65 K}[/tex]

[tex]P_{1} = \text{1.20 atm}[/tex]

Unknown:

[tex]T_{1}[/tex]

Solution:

[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]

[tex]T_{1} = T_{2} × \frac{P_{1}}{P_{2}}[/tex]

[tex]T_{1} = \text{313.65 K} × \frac{\text{1.20 atm}}{\text{2.00 atm}}[/tex]

[tex]\boxed{T_{1} = \text{188.2 K}}[/tex]

[tex]\\[/tex]

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