Assuming that the volume is constant. There is 102.4 kPa initially associated with 36 degrees Celsius. What will the final pressure be if temperature increased to 92 degrees Celsius ?
The equations describing these laws are special cases of the ideal gas law, PV = nRT, where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (universal) gas constant.
I find it too long and too cumbersom to go through the ideal gas law. Faster to use
(P1/T1) = (P2/T2). Remember to convert T to kelvin. Post your work if you get stuck.
To solve this question, we can use the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
where:
P1 and P2 are the initial and final pressures respectively,
V1 and V2 are the initial and final volumes respectively, and
T1 and T2 are the initial and final temperatures respectively.
In this case, we are assuming that the volume is constant, which means V1 = V2. Therefore, the equation can be simplified to:
P1 / T1 = P2 / T2
Let's plug in the given values:
P1 = 102.4 kPa (initial pressure)
T1 = 36°C + 273.15 = 309.15 K (initial temperature in Kelvin)
T2 = 92°C + 273.15 = 365.15 K (final temperature in Kelvin)
Now we can rearrange the equation to solve for P2:
P2 = (P1 * T2) / T1
P2 = (102.4 kPa * 365.15 K) / 309.15 K
Calculating this, we find:
P2 ≈ 121.30 kPa
Therefore, the final pressure will be approximately 121.30 kPa when the temperature increases to 92 degrees Celsius, assuming the volume remains constant.