Question

A gas has a pressure of 0.40 atm at 55.0 °C. What is the pressure at standard temperature?

please show the formula and which law is it?! thanks!!

Answer 1

Gay-Lussac's Law; p₁/T₁ = p₂/T₂; 0.33 atm

Step-by-step explanation:

The volume and number of moles are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

p₁/T₁ = p₂/T₂ Invert each side of the equation

T₁/p₁ = T₂/p₂ Multiply each side by p₂

T₂ = T₁ × p₁/p₂

p₂ = p₁ × T₂/T₁

Data:

p₁ = 0.40 atm; T₁ = 55.0 °C

p₂ = ?; T₂ = 0.0 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = 55.0 + 273.15 = 328.15 K

T₂ = 0.0 + 273.15 = 273.15 K

(b) Calculate the new pressure

p₂ = 0.40 × 273.15/328.15

= 0.40 × 0.832

= 0.33 atm

The pressure drops to 0.33 atm.

Answer 2

0.33 atm.

Step-by-step explanation

By the Ideal Gas Law:

`P \cdot V = n \cdot R\cdot T`

Where

• 
  `P` is the pressure of this gas,
• 
  `V` is the volume of the ideal gas,
• 
  `T` is the temperature of the gas in degrees Kelvins.
• 
  `n` is the number of moles of gas particles in this gas, and
• 
  `R` is the ideal gas constant.

`T_1 =55.0 \; \textdegree\text{C} = (55.0 + 273.15) = 328.15 \; \text{K}`.

`T_2 = 0 \; \textdegree\text{C} = 273.15 \; \text{K}` under STP standard conditions.

`P_1 = (n \cdot R\cdot T_1)/(V)`,

`P_2 = (n \cdot R\cdot T_2)/(V)`.

The value of
`n`,
`R`, and
`V` shall be the same in the two equations.

Divide the second equation with the first:

`(P_2)/(P_1) = (n\cdot R\cdot (1)/(V))/(n\cdot R\cdot (1)/(V)) \cdot (T_2)/(T_1) = (T_2)/(T_1)`.

`P_2 = P_1 \cdot (T_2)/(T_1) = 0.40 * (273.15)/(328.15) = 0.33 \; \text{atm}`.