Question

A gas that has a volume of 28 L , a temperature 45 Celsius and an unknown pressure has its volume increase a 34 L in the temperature decrease to 35 Celsius if I measure the pressure after the change to be 2.0 ATM what was the original pressure of the gas

Answer 1

2.5 atm

Step-by-step explanation:

To find the original pressure of the gas, you need to use the Combined Gas Law. The corresponding equation is:

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

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

Before you can plug the given values into the equation and solve for "P₁", you need to convert the temperatures from Celsius to Kelvin (°C + 273 = K).

The final answer should have 2 sig figs like the least accurate given value.

P₁ = ? atm P₂ = 2.0 atm

V₁ = 28 L V₂ = 34 L

T₁ = 45°C + 273 = 318 K T₂ = 35°C + 273 = 308 K

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

`(P_1(28L))/(318K)=((2.0 atm)(34L))/(308K)` <----- Insert values

`(P_1(28L))/(318K)=0.2208(atm*L)/(K)` <----- Simplify right side

`{P_1(28L)}=70.21atm*L` <----- Multiply both sides by 318 K

`P_1=2.5atm` <----- Divide both sides by 28 L