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?

Answer 1

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.

Answer 2

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.

Answer 3

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!