When the temperature of 4 dm'3 of an ideal gas is increased from 200k to halved, calculate the final volume of the gas.
what does "increased from 200k to halved" even mean?
In any case, PV = kT
To solve this problem, we can use the ideal gas law, which states:
PV = nRT
Where:
P = Pressure of the gas
V = Volume of the gas
n = Number of moles of gas
R = Ideal gas constant
T = Temperature of the gas
In this case, the initial volume (V1) is 4 dm^3, the initial temperature (T1) is 200 K, and the final temperature (T2) is halved:
V1 = 4 dm^3
T1 = 200 K
T2 = T1/2 = 200 K / 2 = 100 K
Next, we need to use the ideal gas law to calculate the final volume (V2). To do this, we'll set up the equation with the known values and solve for V2:
(PV1) / T1 = (PV2) / T2
Since we're assuming the gas is ideal, we can cancel out the pressure (P) and the number of moles (n) since they remain constant. Rearranging the equation, we get:
V2 = V1 * (T2 / T1)
Now, let's substitute the values into the equation:
V2 = 4 dm^3 * (100 K / 200 K) = 4 dm^3 * 0.5 = 2 dm^3
Therefore, the final volume of the gas is 2 dm^3.
To calculate the final volume of the gas when the temperature is halved, we can use the relationship between temperature and volume for an ideal gas, known as Charles' Law.
Charles' Law states that, at constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature. Mathematically, this can be expressed as:
V₁ / T₁ = V₂ / T₂
Where:
- V₁ is the initial volume of the gas
- T₁ is the initial temperature of the gas
- V₂ is the final volume of the gas
- T₂ is the final temperature of the gas
In this case, we are given that the initial volume (V₁) is 4 dm³ and the initial temperature (T₁) is 200 K. We need to calculate the final volume (V₂) when the temperature is halved, which means T₂ = T₁ / 2.
Substituting the values into the equation, we have:
4 dm³ / 200 K = V₂ / (200 K / 2)
Simplifying the right side:
4 dm³ / 200 K = V₂ / 100 K
To isolate V₂, we can cross-multiply:
V₂ = (4 dm³ / 200 K) * 100 K
V₂ = 2 dm³
Therefore, the final volume of the gas is 2 dm³ when the temperature is halved.