Question

A sample of helium gas has a volume of 1.50 L at 159 K and 5.00 atm. When the gas is compressed to 0.200 L at 50.0 atm, the temperature increases markedly. What is the final temperature? Enter your answer in the provided box. K

Answer 1

Answer:

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Explanation:

The given data is as follows.

     $V_{1}$ = 1.50 L,    $T_{1}$ = 159 K,

      $P_{1}$ = 5.00 atm,   $V_{2}$ = 0.2 L,

       $T_{2}$ = ?,       $P_{2}$ = 50.0 atm  

And, according to ideal gas equation,  

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

Hence, putting the given values into the above formula to calculate the value of final temperature as follows.          

 $\frac{5.0 atm \times 1.50 L}{159 K} = \frac{5 atm \times 0.2 L}{T_{2}}$

            $T_{2}$ = 21.27 K

Thus, we can conclude that the final temperature is 21.27 K .