The volume of gas varies directly with temperature and inversely with pressure. Volume is 100 m^3 when temperature is 1508 degrees and pressure is 15 lb/cm^2. What is the volume when the temperature is 2508 degrees and the pressure is 20 lb/cm^2?
To find the volume when the temperature is 2508 degrees and the pressure is 20 lb/cm^2, we can use the concept of direct and inverse variation.
Let's define the variables:
V₁ = Initial volume (100 m³)
T₁ = Initial temperature (1508 degrees)
P₁ = Initial pressure (15 lb/cm²)
V₂ = Final volume (unknown)
T₂ = Final temperature (2508 degrees)
P₂ = Final pressure (20 lb/cm²)
According to the given information, we have the following equations:
Direct Variation:
V₁ / T₁ = V₂ / T₂ [Volume varies directly with temperature]
Inverse Variation:
V₁ * P₁ = V₂ * P₂ [Volume varies inversely with pressure]
Now, we can substitute the given values into the equations and solve for V₂:
Using direct variation, we get:
V₁ / T₁ = V₂ / T₂
100 / 1508 = V₂ / 2508 [Substituting values]
Now, let's solve for V₂:
V₂ = (100 / 1508) * 2508
V₂ ≈ 166.02 m³
Using inverse variation, we get:
V₁ * P₁ = V₂ * P₂
100 * 15 = V₂ * 20 [Substituting values]
Now, let's solve for V₂:
V₂ = (100 * 15) / 20
V₂ = 75 m³
Therefore, the volume when the temperature is 2508 degrees and the pressure is 20 lb/cm² is approximately 75 m³.
To solve this problem, we can use the formula for direct and inverse variation:
direct variation: V = k * T
inverse variation: V = k / P
Where:
V is the volume of the gas
T is the temperature
P is the pressure
k is a constant
We can determine the value of k by plugging in the given values for the initial conditions. The volume (V) is 100 m^3, the temperature (T) is 1508 degrees, and the pressure (P) is 15 lb/cm^2:
100 = k * 1508 (direct variation equation)
100 = k / 15 (inverse variation equation)
To solve the first equation, we divide both sides by 1508:
k = 100 / 1508
k ≈ 0.0662
Now, we can use this value of k to find the volume (V) for the new conditions. The temperature (T) is 2508 degrees, and the pressure (P) is 20 lb/cm^2.
For direct variation:
V = k * T
V = 0.0662 * 2508
V ≈ 165.6156 m^3
For inverse variation:
V = k / P
V = 0.0662 / 20
V ≈ 0.00331 m^3
Therefore, the volume of the gas when the temperature is 2508 degrees and the pressure is 20 lb/cm^2 is approximately 165.616 m^3 (using direct variation) or approximately 0.00331 m^3 (using inverse variation).
v = kt/p , where k is a constant
using the given data
100 = k(1508/15
k = 1500/1508
so when t=2508 and p = 20
v = (1500/1508)(2508/20) = appr 124.73 m^3