If You transfer a sample of a gas at 17 °C from a volume of 5.67 L and 1.10 atm to a container at 37 °C that has a pressure of 1.10 atm. What is the new volume of the gas?
Omg
5.5L of a gas at STP is depressurized to 0.5 atmospheres and heated to 303K. What will be its new volume?
To find the new volume of the gas, we can use the combined gas law equation, which relates the pressure, volume, and temperature of a gas.
The combined gas law equation is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Given:
P1 = 1.10 atm (initial pressure)
V1 = 5.67 L (initial volume)
T1 = 17 °C (initial temperature)
P2 = 1.10 atm (final pressure)
T2 = 37 °C (final temperature)
Now, we need to convert the temperatures to Kelvin scale, as the gas law requires temperature to be in Kelvin.
To convert from Celsius to Kelvin, we use the formula: Kelvin = Celsius + 273.15
Temperature conversion:
T1 = 17 °C + 273.15 = 290.15 K
T2 = 37 °C + 273.15 = 310.15 K
Now, we can substitute the values into the combined gas law equation and calculate the new volume (V2):
(1.10 atm * 5.67 L) / (290.15 K) = (1.10 atm * V2) / (310.15 K)
Simplifying the equation, we get:
V2 = (1.10 atm * 5.67 L * 310.15 K) / (1.10 atm * 290.15 K)
The 'atm' units cancel out, as well as the temperature units, leaving us with:
V2 = (5.67 L * 310.15 K) / 290.15 K
Calculating the values:
V2 = (1,757.1505 L*K) / 290.15 K
V2 ≈ 6.06 L
Therefore, the new volume of the gas is approximately 6.06 L when transferred to a container at 37 °C with a pressure of 1.10 atm.
(P1V1)/T1 = (P2V2)/T2
Con't forget to use T1 and T2 in Kelvin.
Kelvin = 273 + degrees C.