A gas sample occupies a volume of 400.0ml at 298.15k at costant pressure. what is the final temperature when the volume occupied decreased by 39.0
A gas sample occupies a volume of 400.0ml at 298.15k at costant pressure. what is the final temperature when the volume occupied decreased by 39.0?
PV=kT
Since P is constant,
V/T = k/P is constant
if V is decreased by a factor of 361/400, T must decrease by the same factor to have the quotient remain constant.
298.15 * 361/400 = 269.08K
It seems odd that the temperature goes down as the gas is squeezed, but remember that we are keeping a constant pressure.
To find the final temperature, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
In this case, we are given the initial volume (400.0 ml), the initial temperature (298.15 K), and the change in volume (-39.0 ml). We also know that the pressure remains constant.
To solve for the final temperature, we need to find the final volume and then substitute the values into the equation.
The final volume can be calculated using the formula:
Final Volume = Initial Volume + Change in Volume
Final Volume = 400.0 ml - 39.0 ml
Final Volume = 361.0 ml
Now, we can substitute the known values into the ideal gas law equation to solve for the final temperature. Rearranging the equation to solve for T:
T = (PV) / (nR)
Since the pressure, number of moles, and the gas constant remain constant, we can simplify the equation to:
T_final = (V_final * T_initial) / V_initial
Substituting the values:
T_final = (361.0 ml * 298.15 K) / 400.0 ml
Calculating the final temperature:
T_final = 270.94 K
Therefore, the final temperature when the volume decreases by 39.0 ml is approximately 270.94 K.