the volume of a gas varies inversely as its pressure and directly as its temperature.
A certian gas occupies a volume of 2.3L at a temp of 380 K and a pressure of 20 N/cm^2
find the volume when the temp is 342 K and the pressure is 10 N/cm^2
since v = kT/P,
PV/T is a constant. So,
342*V/10 = 380*2.3/20
To solve this problem, we need to use the inverse and direct variation relationships given in the problem.
Inverse variation means that if one variable increases, the other variable will decrease proportionally. In this case, as the pressure decreases, the volume will increase.
Direct variation means that if one variable increases, the other variable will also increase proportionally. In this case, as the temperature increases, the volume will increase.
Let's assign variables to the given values:
V1 = 2.3L (initial volume)
T1 = 380K (initial temperature)
P1 = 20 N/cm^2 (initial pressure)
We can use the following formula to express the relationship:
V1 ∝ T1P1
Now let's find the constant of variation (k) by substituting the initial values:
k = V1 / (T1 * P1)
Now, we need to find the new volume (V2) when the temperature (T2) is 342K and the pressure (P2) is 10 N/cm^2.
Using the formula:
V2 = k * T2 * P2
Let's substitute the values:
V2 = (V1 / (T1 * P1)) * T2 * P2
Now we can calculate the volume:
V2 = (2.3L / (380K * 20 N/cm^2)) * 342K * 10 N/cm^2
Simplifying the equation:
V2 = (2.3L / 7600 N*cm/K) * (342 K * 10 N/cm^2)
Calculating:
V2 = 0.0003026316 * 3420 L*N
V2 = 1.036 L
Therefore, when the temperature is 342K and the pressure is 10 N/cm^2, the volume of the gas will be approximately 1.036L.