Question

An increase in temperature the kinetic energy and average speed of the gas particles. As a result, the pressure on the walls of the container . Answer Bank What temperature must a gas, initially at 10 ∘C, be brought to for the pressure to triple?

Answer 1

Answer:

a

The pressure will increase

b

$T_2 = 576^oC$

Explanation:

From the ideal gas law we have that

     $PV = nRT$

We see that the temperature varies directly with the pressure so if there is an increase in temperature that pressure will increase

   The initial  temperature is $T_i = 10^oC = 10 + 273 = 283 \ K$

The objective of this solution is to obtain the temperature of the gas where the pressure is tripled

Now from the above equation given that nR and V  are constant  we have that

    $\frac{P}{T} = constant$

=>  $\frac{P_1}{T_1} =\frac{P_2}{T_2}$

Let assume the initial  pressure is $P_1 = 1 Pa$

So tripling it will result  to the pressure being $P_2 = 3 Pa$

So

     $\frac{1}{283} =\frac{3}{T_2}$  

=>   $T_2 = 3 * 283$

=>    $T_2 = 3 * 283$

=>    $T_2 = 849 \ K$

Converting back to $^oC$

   $T_2 = 849 - 273$

=>  $T_2 = 576^oC$