# 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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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.**

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.

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.