Determine which law is appropriate for solving the following problem. What temperature will 215 mL of a gas at 20 C and 1 atm pressure attain when it is subject to 15 atm of pressure?

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Answer 1 · 2017-06-04T08:46:33

$300^{\circ}$ $300^(@)$ $C$ $"C"$

We must use the "combined gas law" $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $frac(P_(1) V_(1))(T_(1)) = frac(P_(2) V_(2))(T_(2))$ .

In this case, the volume is kept constant, so $P_{1} = P_{2}$ $P_(1) = P_(2)$ .

We can therefore cancel the terms from the equation:

$\Rightarrow \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $Rightarrow frac(V_(1))(T_(1)) = frac(V_(2))(T_(2))$

$\Rightarrow \frac{1 atm}{20^{\circ} C} = \frac{15 atm}{T_{2}}$ $Rightarrow frac(1 " atm")(20^(@) " C") = frac(15 " atm")(T_(2))$

Let's divide both sides of the equation by $15$ $15$ $atm$ $"atm"$ :

$\Rightarrow \frac{1}{300^{\circ} C} = \frac{1}{T_{2}}$ $Rightarrow frac(1)(300^(@) " C") = frac(1)(T_(2))$

$\Rightarrow {(\frac{1}{300^{\circ} C})}^{- 1} = {(\frac{1}{T_{2}})}^{- 1}$ $Rightarrow (frac(1)(300^(@) " C"))^(- 1) = (frac(1)(T_(2)))^(- 1)$

$\Rightarrow 300^{\circ}$ $Rightarrow 300^(@)$ $C = T_{2}$ $"C" = T_(2)$

$therefore T_(2) = 300^(@)$ $"C"$

Therefore, the gas will attain a temperature of $300^(@)$ $"C"$ when it is subject to $15$ $"atm"$ of pressure.