A 30.0-L sample of methane gas (CH4) has a temperature of 5.0°C if the gas is kept constant, what would the gas volume be if the temperature is lowered to -3.0°C?
For a certain gas sample obeying Charle's law the proportionality constant k has a value of 0.297 L/K, What would the temperature be in °C for a gas volume of 750.0mL?
1. If the gas is kept constant WHAT. I assume you mean constant pressure.
30.0 L = V1
5.0 C = (273 + 5.0 kelvin) =T1
final volume = V2 in L
final T = 273 + C = 273 + (-3) = 270 K = T2
Use V1/T1 = V2/T2.Substitute and solve for V2 in L.
Post your work if you get stuck.
2. V/degrees K = k = 0.279
Plug in 750 for V and solve fir degreeK, then convert to degrees C. Post youw work if you get stuck.
I got a little question in #2 the answer will be negative degrees? thanks for the help!
No, I don't get a negative number for degrees C. Post your work if you want me to check it.
To solve both of these problems, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature, assuming the pressure and amount of gas are constant. Mathematically, we can express Charles's Law as:
V1 / T1 = V2 / T2
where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
Let's start with the first problem:
We are given:
V1 = 30.0 L (initial volume)
T1 = 5.0°C (initial temperature)
T2 = -3.0°C (final temperature)
We are asked to find V2. Rearranging the equation, we have:
V2 = (V1 * T2) / T1
Substituting the known values:
V2 = (30.0 L * -3.0°C) / 5.0°C
Now we can calculate the final volume:
V2 = -18.0 L°C / °C = -18.0 L (since the units of °C cancel out)
However, volume cannot be negative, so we can disregard the negative sign, and the final volume would be 18.0 L.
Now, moving on to the second problem:
We are given:
V1 = 750.0 mL (initial volume)
T1 = ? (initial temperature)
V2 = ? (final volume)
T2 = ? (final temperature)
k = 0.297 L/K (proportionality constant)
We are asked to find T1, the initial temperature. Rearranging Charles's Law equation, we have:
T1 = (V1 * T2) / (V2 * k)
Substituting the known values:
T1 = (750.0 mL * T2) / (V2 * k)
Note that we need to convert the initial volume from mL to L, since the constant k is expressed in L.
V1 = 750.0 mL = 750.0 mL / 1000 mL/L = 0.750 L
Now we can calculate the initial temperature:
T1 = (0.750 L * T2) / (V2 * k)
Remember to substitute the known values of V2 (final volume) and k (proportionality constant) into the equation to calculate T1.