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.

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.

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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.

Approved by eNotes Editorial Team

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.

Approved by eNotes Editorial Team