If I initially have a gas with pressure of 84kPa and temp of 35 deg C and I heat it an additional 230 deg, what will the new pressure be?  Assume the volume of container is constant.  

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lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

The given in the problem are:

P_1 = 84 kPa   and   T_1 = 35 deg. C

Since 230 deg. C is added, the new temperature becomes:

T_2=35 + 230 = 265 deg. C

To solve for P_2, apply Gay-Lussac's Law since volume is constant.
The formula is:

`P_1/T_1=P_2/T_2`

`84/35=P_2/265`

To get P_2 only at the right, multiply both sides by 265.

`265* 84/35= P_2/265*265`


`636=P_2`

Hence, the new pressure is 636 kPa.

lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

Sorry for the slip in the solution above.

You are right. When applying the formula of Gay-Lussac's Law, the temperatures are measured in Kelvin.

Hence, T_1 and T_2 should be converted from Celsius to Kelvin.

mpumpkin's profile pic

mpumpkin | (Level 1) Valedictorian

Posted on

The given in the problem are:

P_1 = 84 kPa   and   T_1 = 35 deg. C

Since 230 deg. C is added, the new temperature becomes:

T_2=35 + 230 = 265 deg. C

To solve for P_2, apply Gay-Lussac's Law since volume is constant.
The formula is:





To get P_2 only at the right, multiply both sides by 265.




Hence, the new pressure is 636 kPa.

Would I not need to change to Kelvins which makes it 146.72 kPa in the answer?  I am confused about this subject.  

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