A sample of gas at 3.00 x 103 atm inside a steel tank is cooled from -200 °C to 27 °C. What is the final pressure in mmHg of the gas in the steel tank?
We can use the combined gas law to solve for the final pressure:
(P1V1/T1) = (P2V2/T2)
We know that P1 = 3.00 x 103 atm, T1 = -200 °C + 273.15 = 73.15 K, V1 is unknown, T2 = 27 °C + 273.15 = 300.15 K, and we want to solve for P2 in mmHg.
We can convert atm to mmHg by multiplying by 760 mmHg/1 atm, and we can use the ideal gas law to solve for V1:
PV = nRT
V1 = nRT1/P1
where n is the number of moles of gas, R is the gas constant, and T1 is in Kelvin.
Assuming the number of moles and gas constant are constant, we can write:
(P1V1/T1) = (P2V2/T2)
(P2)(V2) = (P1)(V1)(T2/T1)
(P2)(V2) = (3.00 x 103 atm)(nRT1/P1)(300.15 K/73.15 K)
(P2)(V2) = (3.00 x 103)(nRT1)(4.1)
(P2)(V2) = (12.3 x 103)(nRT1)
V2 = (12.3 x 103)(nRT1)/(P2)
Now we can substitute this expression for V2 into our equation and solve for P2:
(P2)(12.3 x 103)(nRT1)/(P2) = (12.3 x 103)(nRT1)
P2 = (12.3 x 103)(nRT1)/[(12.3 x 103)(nRT1)/P1*(T2/T1)]
P2 = P1(T2/T1)
P2 = (3.00 x 103 atm)(300.15 K/73.15 K)
P2 = 1.23 x 104 atm
Finally, we can convert this to mmHg by multiplying by 760 mmHg/1 atm:
P2 = 1.23 x 104 atm x 760 mmHg/1 atm
P2 = 9.36 x 106 mmHg
Therefore, the final pressure of the gas in the steel tank is 9.36 x 106 mmHg.
Since the volume is constant, P/T is constant. So you want P such that
P/(20+273) = 3.0*10^3/(-200+273)
Now just find P.
You are correct that the pressure-temperature relationship for a fixed volume of gas can also be described by the formula P/T = constant.
Using this formula and the given initial and final temperatures, we can set up the following equation to solve for the final pressure in units of atm:
P1/T1 = P2/T2
P2 = P1(T2/T1) = (3.00 x 103 atm)(300.15 K/73.15 K) = 1.23 x 104 atm
To convert this to mmHg, we can multiply by 760 mmHg/1 atm to get:
P2 = 1.23 x 104 atm x 760 mmHg/1 atm = 9.36 x 106 mmHg
Therefore, the final pressure of the gas in the steel tank is 9.36 x 106 mmHg. Thank you for your input!
To solve this problem, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law can be written as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Given:
P1 = 3.00 x 10^3 atm
T1 = -200 °C = -200 + 273.15 = 73.15 K
T2 = 27 °C = 27 + 273.15 = 300.15 K
Now, we need to convert the pressures to the same unit. 1 atm = 760 mmHg, so
P1 = 3.00 x 10^3 atm * 760 mmHg/atm = 2.28 x 10^6 mmHg
Now, we can plug the values into the combined gas law and solve for P2:
(2.28 x 10^6 mmHg * V1) / (73.15 K) = (P2 * V2) / (300.15 K)
Since both Volumes V1 and V2 are unknown, we can assume that they remain constant. Therefore, V1/V2 = 1.
(2.28 x 10^6 mmHg * 1) / (73.15 K) = (P2 * 1) / (300.15 K)
2.28 / 73.15 = P2 / 300.15
Cross-multiplying, we get:
(2.28 * 300.15) / 73.15 = P2
P2 ≈ 9.32 mmHg
Therefore, the final pressure in mmHg of the gas in the steel tank is approximately 9.32 mmHg.