what is the starting temperatue of 150ml of gas when cooled to 33 degrees of celsius and a volume of 120ml?
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To determine the starting temperature of the gas, we can use the equation of state for gases:
(V1/T1) = (V2/T2)
Where:
V1 = initial volume of the gas (150 mL)
T1 = initial temperature of the gas (unknown)
V2 = final volume of the gas (120 mL)
T2 = final temperature of the gas (33 degrees Celsius)
Let's substitute the given values into the equation and solve for T1:
(150 mL / T1) = (120 mL / 33 degrees Celsius)
Now, let's cross-multiply and solve for T1:
150 mL * 33 degrees Celsius = 120 mL * T1
4,950 mL·degrees Celsius = 120 mL * T1
T1 = (4,950 mL·degrees Celsius) / 120 mL
T1 ≈ 41.25 degrees Celsius
Therefore, the starting temperature of the gas is approximately 41.25 degrees Celsius.
To determine the starting temperature of the gas, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant
T = temperature of the gas in Kelvin
Since we are given the initial volume (V1 = 150 mL) and final volume (V2 = 120 mL) of the gas, we can determine the ratio of the volumes (V2/V1). This ratio would be used to find the initial temperature (T1) in terms of the final temperature (T2).
Now, let's calculate the ratio of volumes:
V2/V1 = 120mL / 150mL
= 0.8
The temperature (T) should be in Kelvin, so we need to convert the given Celsius temperature (33 degrees) to Kelvin by adding 273.15:
T2 = 33 + 273.15
= 306.15 K
Now, we can rearrange the ideal gas law equation to solve for T1:
(V2/V1) = (T2/T1)
Substituting the values we have:
0.8 = (306.15 K / T1)
To isolate T1, we can rearrange the equation:
T1 = (306.15 K) / 0.8
≈ 382.69 K
Therefore, the starting temperature of the gas, when cooled to 33 degrees Celsius (306.15 K) and a volume of 120 mL, is approximately 382.69 Kelvin.
at constant pressure?
V1/T1=V1/T2 Charles' law
150ml/T1=120ml/((273+33)
solve for T1