The volume of a gas varies vvaries inversely as the pressure p and directly as the temperature t.
a) Write an equation to represent the volume of a gas in terms of pressure and temperature.
b) Is your equation a direct, joint, inverse, or combined variation?
c) A certain gas has a volume of 8 liters, a temperature of 275 Kelvin, and a pressure of 1.25 atmospheres. If the gas is compressed to a volume of 6 liters and is heated to 300 Kelvin, what will the pressure be? (show your complete solution)
a) The equation to represent the volume of a gas in terms of pressure (p) and temperature (t) is:
V = k * (t/p)
where V is the volume of the gas, k is the constant of variation.
b) This equation represents combined variation as it has both direct and inverse variation.
c) To find the pressure when the gas is compressed to a volume of 6 liters and heated to 300 Kelvin, we can use the equation of combined variation:
V1 * t1 / p1 = V2 * t2 / p2
Let's substitute the given values into the equation:
(8 L * 275 K) / (1.25 atm) = (6 L * 300 K) / p2
Simplifying the equation:
(2200 L*K) / (1.25 atm) = (1800 L*K) / p2
Cross-multiplying:
2200 L*K * p2 = 1.25 atm * 1800 L*K
Dividing both sides by 2200 L*K:
p2 = (1.25 atm * 1800 L*K) / (2200 L*K)
p2 = 1.0227 atm
Therefore, when the gas is compressed to a volume of 6 liters and heated to 300 Kelvin, the pressure will be approximately 1.0227 atmospheres.
a) To represent the volume of a gas in terms of pressure and temperature, we can use the following equation:
V = k * (T / P)
Where V is the volume of the gas, T is the temperature, P is the pressure, and k is the constant of variation.
b) In this equation, the volume V varies inversely with the pressure P, and directly with the temperature T. Therefore, it is an example of combined variation because it combines both direct and inverse variation.
c) To find out the pressure when the volume is 6 liters and the temperature is 300 Kelvin, we can use the given information of the gas having a volume of 8 liters, a temperature of 275 Kelvin, and a pressure of 1.25 atmospheres.
First, we can find the value of the constant of variation k by substituting the initial values into the equation:
8 = k * (275 / 1.25)
Simplifying this equation, we get:
8 = k * 220
Now, we can solve for k:
k = 8 / 220
k = 0.0364
Next, we can use this value of k to find the pressure P when the volume is 6 liters and the temperature is 300 Kelvin:
6 = 0.0364 * (300 / P)
To isolate P, we can rearrange the equation:
6 * P = 0.0364 * 300
Simplifying further:
6P = 10.92
Dividing both sides by 6, we get:
P = 10.92 / 6
P ≈ 1.82 atmospheres
Therefore, when the gas is compressed to a volume of 6 liters and heated to 300 Kelvin, the pressure will be approximately 1.82 atmospheres.
V = k(t/p) , where k is a constant ,
for your data,
8 = k(275/1.25)
k = 2/55
V = (2/55)(t/p)
when V = 6, t = 300, p = ?
sub your values into the equation to find p