A rigid container is filled with chlorine gas. The gas has a pressure of 2.75 bar. The tank is then cooled down to -20.0oC at wh
ich point the pressure of the gas is now measured to be 1.48 bar. What was the original temperature (in oC) of the gas?
1 answer:
Answer:
Original temperature (T1) = - 37.16°C
Explanation:
Given:
Gas pressure (P1) = 2.75 bar
Temperature (T2) = - 20°C
Gas pressure (P2) = 1.48 bar
Find:
Original temperature (T1)
Computation:
Using Gay-Lussac's Law
⇒ P1 / T1 = P2 / T2
⇒ 2.75 / T1 = 1.48 / (-20)
⇒ T1 = (2.75)(-20) / 1.48
⇒ T1 = -55 / 1.48
⇒ T1 = - 37.16°C
Original temperature (T1) = - 37.16°C
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