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GoldenToad7449
13.01.2020 •
Chemistry
The starting pressure of a 15°c system was 1.50 atm. if the pressure rose to 1216 torr, what is the new temperature in celsius?
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Ответ:
The new temperature in Celsius is 34.2°
Explanation:
Let's apply the Charles Gay Lussac law to solve this, where pressure varies directly proportional to Absolute Temperature.
P1 / T1 = P2/T2
Volume and number of moles are still the same in both situations
We must convert T° C to K
15°C + 273 = 288K
And we must convert Torr to atm
760 Torr 1 atm
1216 Torr (1216 .1)/760 = 1.6
1.50atm/288K = 1.60atm/T2
T2 = 1.60 atm .288K/1.5atm
T2 = 307.2K
307.2K - 273 = 34.2°C
Ответ:
The new temperature is 34.21 °C
Explanation:
Step 1: Data given
Temperature of the system = 15°C
Pressure at 15°C is 1.50 atm
The pressure rose to 1216 torr (=1216/760 = 1.6 atm)
Step 2: Calculate the new temperature
P1/T1 = P2/T2
⇒ with P1 = the initial pressure = 1.50 atm
⇒ with T1 = the initial temperature = 15°C = 288.15 K
⇒ with P2 = the initial pressure = 1.60 atm
⇒ with T2 = the initial temperature = TO BE DETERMINED
T2 = T1 * (P2 / P1)
T2 = 288.15 * (1.60/1.50)
T2 = 307.36 K = 34.21 °C
The new temperature is 34.21 °C
Ответ:
Therefore, when an atom becomes a positive ion is pulls its electrons closer, decreasing is radius moreover, when it becomes a negative ion, it pulls its electrons closer and decreases the radius.