Question

a sample of nitrogen gas occupies a volume of 2.00 L at 756 mm Hg and 0.00 C. the volume increases by 2.00 L and the temperature decreases to 137 K. what is final pressure exerted on the gas?

Answer 1

P₂ = 0.250 atm or 190. mmHg**

Step-by-step explanation:

To find the final pressure of the gas, you need to use the Combined Gas Law:

`(P_1V_1)/(T_1) = (P_2V_2)/(T_2)`

In this equation, "P₁", "V₁", and "T₁" represent the initial pressure, volume, and temperature. "P₂", "V₂", and "T₂" represent the final pressure, volume, and temperature.

Before you can solve for "P₂", you need to...

(1) convert the pressure from mmHg to atm (760 mmHg = 1 atm)

(2) convert the temperature from Celsius to Kelvin (°C + 273 = K)

The final answer should have 3 sig figs like the other given values.

P₁ = 756 mmHg / 760 = 0.995 atm P₂ = ? atm

V₁ = 2.00 L V₂ = 2.00 L + 2.00 L = 4.00 L

T₁ = 0.00°C + 273 = 273 K T₂ = 137 K

`(P_1V_1)/(T_1) = (P_2V_2)/(T_2)` <----- Combined Gas Law

`((0.995atm)(2.00L))/(273K) = (P_2(4.00L))/(137K)` <----- Insert values

`0.007287(atm*L)/(K) = (P_2(4.00L))/(137K)` <----- Simplify left side

`0.998atm*L = P_2(4.00L)` <----- Multiply both sides by 137 K

`0.250atm = P_2` <----- Divide both sides by 4.00 L

** 0.250 atm x 760 = 190. mmHg