The pressure of oxygen gas inside a container is 5.00 atm at 25°C. If the temperature falls to -10°C, what is the new pressure inside the container?

Question

Wyatt Atkins · Accepted Answer

The new pressure is 4.41 atm.

Given The pressure (#P_1#) of a gas at a temperature #T_1# A second temperature #T_2# Find The second pressure #P_2# Strategy A problem involving two gas pressures and two temperatures must be a Gay-Lussac's Law problem. Gay-Lussac's Law is #color(blue)(|bar(ul(color(white)(a/a) P_1/T_1 = P_2/T_2color(white)(a/a)|)))" "# In your problem, #P_1 = "5.00 atm"; T_1 = "(25 + 273.15K)" color(white)(ll)= "298.15 K"# #P_2 = "?"; color(white)(mmmll)T_2 = "(-10 + 273.15 K)" = "263.15 K"# Insert the numbers into the Gay-Lussac's Law expression. Solution #P_1/T_1 = P_2/T_2# #P_2 = P_1 × T_2/T_1 = "5.00 atm" × (263.15 cancel("K"))/(298.15 cancel("K")) = "4.41 atm"# The new pressure is 4.41 atm.

Cooper Adams · Answer

undefined Using the ideal gas law, we can solve for the new pressure:

P1 = 5.00 atm (initial pressure)
T1 = 25°C + 273.15 = 298.15 K (initial temperature)
T2 = -10°C + 273.15 = 263.15 K (final temperature)

P2 = (P1 * T2) / T1
P2 = (5.00 atm * 263.15 K) / 298.15 K
P2 ≈ 4.41 atm

Therefore, the new pressure inside the container is approximately 4.41 atm.