Question

A gas is in a sealed container.Part ABy what factor does the gas temperature change if the volume is doubled and the pressure is tripled?Part BBy what factor does the gas temperature change if the volume is halved and the pressure is tripled?

Answer 1

A) The temperature increases by a factor 6

We can use the ideal gas equation:

$pV=nRT$

where

p is the pressure

V is the volume

n is the number of moles

R is the gas constant

T is the temperature

We can also rewrite it as

$\frac{pV}{T}=nR$

The gas is in a sealed container - this means the amount of gas is fixed, so n is constant. Since R is constant too, the term on the right in the equation is constant. So we can rewrite the equation as:

$\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where in this problem we have:

$V_2 = 2V_1$ (the volume is doubled)

$p_2 = 3 p_1$ (the pressure is tripled)

re-arranging the equation, we find the change in temperature:

$\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(2 V_1)}{p_1 V_1}=6$

so, the temperature increases by a factor 6.

B) The temperature increases by a factor 1.5.

We can use again the same equation:

$\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}$

Where in this case:

$V_2 = \frac{V_1}{2}$ (the volume is halved)

$p_2 = 3 p_1$ (the pressure is tripled)

So, we can find the change in temperature:

$\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(\frac{V_1}{2})}{p_1 V_1}=\frac{3}{2}=1.5$

So, the temperature increases by 1.5 times.