Question

Consider 5.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.30 L and the temperature is increased to 40. ∘C , what is the new pressure, P2, inside the container? Assume no change in the amount of gas inside the cylinder.

Answer 1

Answer:

P₂ = 1.12 atm

Explanation:

To find the new pressure, you need to use the Combined Gas Law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

In this equation, "P₁", "V₁", and "T₁" represent the initial pressure, volume, and temperature. "P₂", "V₂", and "T₂" represent the new pressure, volume, and temperature. Before plugging the values into the equation, you need to

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

(2) convert the temperatures from Celsius to Kelvin (°C + 273)

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

P₁ = 365 mmHg / 760 = 0.480 atm           P₂ = ? atm

V₁ = 5.00 L                                                   V₂ = 2.30 L

T₁ = 20°C + 273 = 293 K                             T₂ = 40°C + 273 = 313 K

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$                                              <----- Combined Gas Law

$\frac{(0.480 atm)(5.00 L)}{293 K}=\frac{P_2(2.30 L)}{313 K}$                       <----- Insert values

$0.00819=\frac{P_2(2.30 L)}{313 K}$                                     <----- Simplify left side

$2.56 = P_2(2.30L)$                                      <----- Multiply both sides by 313

$1.12 = P_2$                                                  <----- Divide both sides by 2.30