Help my fellow classmates with this question please

Question

A sample of refrigeration gas has a volume of 405 mL at a pressure of 1.50 atm and a temperature of 28°C.
The gas was compressed into a volume of 130 mL with a pressure of 3.8 atm. What temperature in degrees Celsius did the gas change to?

Answer 1

See

https://www.wyzant.com/resources/answers/507783/a_sample_of_refrigeration_gas_has_a_volume_of_405_ml_at_a_pressure_of_1_50_atm_and_a_temperature_of_28_176_c

Answer 2

Begin with PV = nRT (always on a gas law problem!)

 

Identify what variables are listed: V, P, and T.

 

So, use these in an equation:

 

(p1 x v1)/t1 = (p2 x v2)/t2

 

Solve for t2  because " What temperature in degrees Celsius did the gas change to?"

 

t2 = (p2 x v2)/(p1 x v1)/t1 = (p2 x v2 x t1)/p1 x v1

 

Convert t1 to Kelvin (all gas law problems in Kelvin - always) t1 = 28C = 273 + 28 = 301 K

 

Substitute in:

 

(3.8 atm x 130 mL x 301 K)/1.50 atm x 405 mL = ? K

 

Convert back to C by subtracting 273.

 

Can you complete this now?

Answer 3

Hello Karen

 

Use the combined gas law equation:

 

P₁ V₁         P₂ V₂
---------   =  ---------

    T₁              T₂

 

P₁ = 1.50 atm                          P₂ = 3.8 atm

V₁ = 405 mL = 0.405 L           V₂ = 130 mL = 0.13 L

T₁ = 28°C + 273 = 301 K       T₂ = ?

 

Rearrange to solve for T₂

 

             P₂ V₂ T₁  

T₂  =   ----------------

                P₁ V₁

 

            3.8 atm x 0.13 L x 301 K

       =  -----------------------------------

             1.50 atm x 0.405 L

 

       = 245 K

 

T in °C = 245 - 273 =  - 28 °C