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?

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

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

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