Question

A sample of argon has a volume of 5.0 dm^3 and the pressure is 0.92 atm.
If the final temperature is 30.° C, the final volume is 5.7 L, and the final
pressure is 800. mm Hg, what was the initial temperature of the argon?

Answer 1

Answer:

232.9K

       

Explanation:

V₁  = 5dm³

P₁  = 0.92atm

T₂ = 30°C

V₂ = 5.7L

P₂ = 800mmHg

Unknown:

T₁ = ?

Solution:

To solve this problem, we are going to apply the combined gas law which is fusion of the avogadro's law, boyle's law and charles's law.

 Mathematically;

                   $\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }$

p, v and T are pressure, volume and temperature respetively

1 and 2 depicts the initial and final states.

We need to convert ;

     temperature to K and pressure into atm

T₂  = 30°C  = 273 + 30 = 303K

P₂  =  800mmHg;

         760mmHg  = 1 atm

           800mmHg  = $\frac{800}{760}$  = 1.05atm

   Input the variables and solve;

      $\frac{0.92 x 5}{T} = \frac{1.05 x 5.7}{303}$

    T  = 232.9K

       

Answer 2

The initial temperature of the argon is 232.9K.

Calculation of the initial temperature:

Since

V₁  = 5dm³

P₁  = 0.92atm

T₂ = 30°C

V₂ = 5.7L

P₂ = 800mmHg

Here we have to applied the gas law

P1V1/T1 = P2V2/T2

Here p, v, and T are pressure, volume and temperature respetively

And, 1 and 2 depict the initial and final states.

Now we need to convert

temperature to K and pressure into atm

So, it be like

T₂  = 30°C  = 273 + 30 = 303K

P₂  =  800mmHg;

        760mmHg  = 1 atm

So,

= 800/760 = 1.05 atm

Now

0.92*5 /T = 1.05*5.7/303

So, the temperature is 232.92K

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