A sample of nitrogen gas occupies 1.55 L at 27oC and 1 atm pressure. What will the volume be at -100oC and the same pressure?
0.894
2.78
To solve this problem, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where:
P1 = Initial pressure (1 atm)
V1 = Initial volume (1.55 L)
T1 = Initial temperature (27°C + 273.15 = 300.15 K)
P2 = Final pressure (1 atm)
V2 = Final volume (to be determined)
T2 = Final temperature (-100°C + 273.15 = 173.15 K)
Now, we can rearrange the equation and solve for V2:
(P1 * V1 * T2) / (T1 * P2) = V2
Substituting the given values into the equation:
(1 atm * 1.55 L * 173.15 K) / (300.15 K * 1 atm) = V2
Calculating the result:
(268.05725) / (300.15) = V2
V2 ≈ 0.893 L (rounded to three decimal places)
Therefore, the volume of the nitrogen gas at -100°C and 1 atm pressure would be approximately 0.893 L.
To find the volume of the nitrogen gas at -100oC and the same pressure, we can use the combined gas law equation.
The combined gas law equation can be written as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 and P2 are the initial and final pressures respectively,
V1 and V2 are the initial and final volumes respectively,
T1 and T2 are the initial and final temperatures respectively.
In this case, we know:
P1 = P2 = 1 atm (pressure remains constant)
V1 = 1.55 L
T1 = 27oC (which needs to be converted to Kelvin)
T2 = -100oC (which also needs to be converted to Kelvin)
To convert temperatures from Celsius to Kelvin, we use the following formula:
K = C + 273.15
Converting the temperatures:
T1 = 27 + 273.15 = 300.15 K
T2 = -100 + 273.15 = 173.15 K
Now we can plug in the values into the combined gas law equation:
(1 atm * 1.55 L) / (300.15 K) = (1 atm * V2) / (173.15 K)
To solve for V2 (the final volume), we can rearrange the equation as follows:
V2 = (1 atm * V1 * 173.15 K) / (1 atm * 300.15 K)
V2 = (1.55 L * 173.15 K) / (300.15 K)
V2 ≈ 0.894 L
Therefore, the volume of nitrogen gas at -100oC and 1 atm pressure would be approximately 0.894 L.