At standard temperature, a gas has a volume
of 345 mL. The temperature is then increased
to 124◦C, and the pressure is held constant.
What is the new volume?
Answer in units of mL
(V1/T1) = (V2/T2)
Remember T must be in kelvin.
Standard T is 273K
40ml
To find the new volume of the gas when the temperature is increased to 124°C while the pressure is held constant, we need to use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin
First, we need to convert the given temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is:
T(K) = T(°C) + 273.15
So, the temperature in Kelvin would be:
T(K) = 124 + 273.15
T(K) = 397.15 K
Next, since the pressure is held constant, we can rearrange the ideal gas law equation to solve for the new volume V2:
V2 = (V1 * T2) / T1
Where:
V1 is the initial volume of the gas (345 mL)
T1 is the initial temperature of the gas in Kelvin (which is also the temperature in Kelvin before the increase)
T2 is the new temperature of the gas in Kelvin
Substituting the given values into the equation:
V2 = (345 mL * 397.15 K) / T1
Remember to convert mL to liters by dividing by 1000, as the ideal gas constant has units of L.mol⁻¹.K⁻¹.
V2 = (0.345 L * 397.15 K) / T1
Now, we need to find the value of T1. We know that the temperature should be the same initially as the final temperature before the increase, which is 124°C. So we convert this temperature to Kelvin using the formula:
T1 = 124 + 273.15
T1 = 397.15 K
Substituting this value into the equation:
V2 = (0.345 L * 397.15 K) / 397.15 K
Finally, we can simplify the equation:
V2 = 0.345 L
Therefore, the new volume of the gas, when the temperature is increased to 124°C while the pressure is held constant, is 0.345 mL.