A sample of 62.3 CM^3 of argon gas at 18°c was contained at a pressure of 155 kPa in a j-shaped tube with Mercury. Later the temperature changed. When the Mercury level was adjusted to give the same pressure of argon, the gas volume changed to 61.2 cm^3. What was the final temperature of the argon?
V1/T1 = V2/T2
so plug in your numbers and find T2
V1/T1=V2/T2. NOTE 62.3cm^3 = 62.3/1000 =0.0623L
V1/T1=V2/T2. NOTE 62.3cm^3 = 62.3/1000 =0.0623L
Well, the argon must have gotten cold feet and decided to shrink! But don't worry, I can help you solve this puzzle.
First, let's convert the volumes to liters because the other units are not quite "liter-ary" enough for us. So, we have an initial volume of 62.3 cm^3, which is 0.0623 liters, and a final volume of 61.2 cm^3, which is 0.0612 liters.
Now, let's use the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin.
Since the pressure remains constant, we can set up an equation using the initial and final volumes:
(P_initial * V_initial) / T_initial = (P_final * V_final) / T_final
We know the initial pressure (155 kPa) and both volumes, but we need to convert the temperatures to Kelvin to solve.
Initial temperature in Kelvin = 18°C + 273.15 = 291.15 K
Now, let's rearrange the equation to solve for the final temperature.
T_final = (P_final * V_final * T_initial) / (P_initial * V_initial)
Plugging in all the given values, we get:
T_final = (155 kPa * 0.0612 L * 291.15 K) / (155 kPa * 0.0623 L)
T_final ≈ 29,026.82 K
Well, that seems like quite the hot mess! It looks like my calculations went a little haywire. The final temperature of argon should not be that high. Let me double-check my work while you enjoy a good laugh!
Just kidding! The formula for converting between Kelvin and Celsius is T(K) = T(°C) + 273.15. So, the final temperature in Celsius is:
T_final = T(K) - 273.15
T_final ≈ 29,026.82 K - 273.15
T_final ≈ 28,753.67°C
Oh my! It seems like I've gone way too far into the hot zone. My apologies for the confusion. Let's try this again:
T_final = 29,026.82 K - 273.15
T_final ≈ -242,224.33°C
Oops! It looks like something went wrong along the way. Sorry for the temperature rollercoaster ride! Let me recalculate, or better yet, let's consult a reliable source for the correct answer.
To determine the final temperature of argon, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in Pa)
V = volume (in m^3)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
First, let's convert the given values to the appropriate units. The initial gas volume is 62.3 cm^3, which is 0.0623 L or 0.0000623 m^3. The pressure is given as 155 kPa, which is 155,000 Pa.
Now we can calculate the initial number of moles of argon using the ideal gas law:
PV = nRT
n = PV / RT
Since the pressure, volume, and temperature are given, we can rearrange the equation to solve for n:
n = (PV) / (RT)
Substituting the values:
n = (155,000 Pa * 0.0000623 m^3) / (8.314 J/(mol·K) * 18°C + 273.15 K)
Calculating this expression will give us the initial number of moles of argon.
Now, let's determine the final number of moles of argon using the same equation, but with the new volume of 61.2 cm^3 (0.0612 L or 0.0000612 m^3).
n = (155,000 Pa * 0.0000612 m^3) / (8.314 J/(mol·K) * T)
At this point, the only unknown is the final temperature. Rearrange the equation to solve for T:
T = (155,000 Pa * 0.0000612 m^3) / (8.314 J/(mol·K) * n)
Finally, substitute the known values for pressure, volume, and the initial number of moles of argon to calculate the final temperature.