A gas dog certain mass occupies volume of 1.2L at 37 degree Celsius and 3.0 atm.At what temperature will the volume and pressure of this gas become one third of their initial values?
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Given,
Volume 1 (V₁) = 1.2 L
Temperature 1 (T₁) = 37°C = 310 K
Pressure 1 (P₁) = 3 atm
Volume 2 (V₂) = (1.2/3) L = 0.4 L
Pressure 2 (P₂) = (3/3) atm = 1 atm
By Ideal Gas Equation,
P₁V₁/T₁ = P₂V₂/T₂
= (3*1.2)/310 = (1*0.4)/T₂
= 3.6/310 = 0.4/T₂
= 3.6T₂ = 124
= T₂ = 34.45 K
or,
= T₂ = (34.45-273)°C
= T₂ = -238.55°C
I hope this helps.
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Answer:yyfhjjgvkgoykvmvmvyryehcncydydyd5yd
Explanation:
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