The gas is in a sealed can is at a pressure of 3.00 atm at 25 degrees Celsius. What would the gas pressure in the can be at 52.0 degrees Celsius.

Answer 1

Since the volume is constant (a sealed can), we can use P1/T1 = P2/T2 and solve for P2:

Recall, T must be in Kelvin

P1 = 3.00 atm

T1 = 25C + 273 = 298K

P2 = ?

T2 = 52C + 273 = 325K

3.00 atm / 298K = P2/325K

P2 = 3.27 atm

Answer 2

Following PV=nRT, we need to do a few things first. 1. We need to convert Celsius to kelvin by adding 273.15 so 25 C becomes 298.15 K and 52 C becomes 325.15 K. Using this we now move on to our universal gas constant R in PV = nRT which is 0.082057 (L*atm*K^-1*mol^-1) the units are important because we want only atm left. For the case of simplicity I will assume n is 1 mol. We then solve for volume with the known pressure,

V = (nRT)/P = (1*.082057*298.15)/3 = 8.15 Liters (I can't fit in the other units for the R const, temperature and Pressure) using this constant volume we then resolve for pressure using the new temperature of 325.15 K

P = (nRT)/V = (1*0.082057*325.15)/(8.15) = 3.27 atm which is our final answer. The main thing to follow is the solving for your unknown and the fact that the R gas constant will cancel out unwanted units if done correctly.