0.75 mol of argon gas is admitted to an evacuated 40 cm^{3} container at 40 C. The gas then undergoes an isochoric heating to a temperature of 500 C.
What is the question?
I had to look up "isochoric". It means "contant volume". During that heating process, p will increase such that it remains proportional to (absolute) T. The absolute temperature increases by a factor 773/313 = 2.47. Pressure increases by the same factor. They do not tell you the initial pressure, but you can get the pressure using
P = nRT/V
where R = 82.06 cm^3*atm/mole K
Po = 0.75*82.06*313/40 = 482 atm
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Step 1: Calculate the initial pressure
Since the container is initially evacuated, the pressure is zero.
P1 = 0
Step 2: Calculate the final pressure
Since the volume remains constant (isochoric process), the pressure can be calculated using the initial pressure and temperature, and then converted to kelvin.
T1 = 40 °C = 40 + 273.15 = 313.15 K (initial temperature)
T2 = 500 °C = 500 + 273.15 = 773.15 K (final temperature)
P2 = (P1 * T2) / T1
P2 = (0 * 773.15) / 313.15
P2 = 0
Step 3: Calculate the number of moles
We are given that there are 0.75 mol of argon gas.
n = 0.75 mol
Step 4: Calculate the volume
Since the pressure remains constant (zero), we can rearrange the ideal gas law equation to solve for volume:
V = (n * R * T) / P
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
V = (0.75 * 0.0821 * 773.15) / 0
V = 0 (since P2 = 0)
Therefore, the volume of the gas remains zero, which means the gas does not expand or change its volume during the isochoric heating process.