Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 ˚C to 30.0 ˚C.

Question

Henry Antrim · Accepted Answer

There is a 0.03 atm increase in pressure. Gay-Lussac's Law, which states that an ideal gas of fixed volume exerts pressure on its sides correspondingly to its temperature, is applied in this problem. #P_1/T_1=P_2/T_2# #P_1# = 1.00 atm; #T_1# = (20.0 + 273.15) K = 293.2 K.
#P_2# = ?; #T_2# = (30.0 + 273.15) K = 303.2 K. We know #P_1#, #T_1#, and #T_2#. Thus, we can calculate #P_2#. #P_2 = P_1 × T_2/T_1# = 1.00 atm × #(303.2" K")/(293.2" K")# = 1.03 atm #ΔP = P_2 – P_1# = (1.03 -1.00) atm = 0.03 atm There is a 0.03 atm increase in pressure. This makes sense: the pressure should rise by roughly 3% (0.03 atm) when the temperature increases by 10 parts in 300 or 3 parts in 100 (3%).

Axel Alvarado · Answer

undefined According to the ideal gas law, $PV = nRT$, when volume is constant, pressure and temperature are directly proportional. So, the pressure change can be calculated using the formula:

$\Delta P = P \times \frac{\Delta T}{T}$

Given:
Initial pressure ($P$) = 1.00 atm
Initial temperature ($T_1$) = 20.0 ˚C = 293.15 K
Final temperature ($T_2$) = 30.0 ˚C = 303.15 K

$\Delta T = T_2 - T_1 = 303.15 K - 293.15 K = 10.00 K$

$\Delta P = 1.00 \, \text{atm} \times \frac{10.00 \, \text{K}}{293.15 \, \text{K}} $

$\Delta P \approx 0.0341 \, \text{atm}$

So, the pressure change when the gas is heated from 20.0 ˚C to 30.0 ˚C is approximately 0.0341 atm.