Question

A gas is in a sealed container. By what factor does the gas temperature change if:

a. The volume is doubled and the pressure is tripled?

b. The volume is halved and the pressure is tripled?

Short answer

Change in Temperature of the gas for the given conditions are

a. The temperature is increased 6 Times.

b. The temperature is increased 1.5 Times.

Step-by-step solution

Concept Formula (For Part a and Part b) 

By using the ideal gas law, a comparison of the same gas with the same number of moles can be represented by the formula,

$\frac{p_2{V}_2}{T_2}=\frac{p_1{V}_1}{T_1}$,

where $p$is Pressure, $V$is Volume and $T$is Temperature.

Finding the temperature change if volume is doubled and pressure is tripled (for Part a)

As per the data that is provided, the Volume is being doubled.

Hence, $V_2=2V_1$

As per the data that is provided, the Pressure is being tripled.

Hence, $p_2=3p_1$.

Thus, change in temperature can be calculated as,

$\begin{array}{rcl}\frac{p_2{V}_2}{T_2}& =& \frac{p_1{V}_1}{T_1}\\ \frac{\left(3p_1\right)\left(2V_1\right)}{T_2}& =& \frac{p_1{V}_1}{T_1}\\ \frac{6}{T_2}& =& \frac{1}{T_1}\\ T_2& =& 6T_1\end{array}$

Hence, the temperature of the gas increases 6 Times.

Finding the temperature change if volume is halved and pressure is tripled (for Part b)

As per the data that is provided, the Volume is being doubled.

Hence, $V_2=0.5V_1$

As per the data that is provided, the Pressure is being tripled.

Hence, $p_2=3p_1$

Thus, change in temperature can be calculated as,

$\begin{array}{rcl}\frac{p_2{V}_2}{T_2}& =& \frac{p_1{V}_1}{T_1}\\ \frac{\left(3p_1\right)(0.5V_1)}{T_2}& =& \frac{p_1{V}_1}{T_1}\\ \frac{1.5}{T_2}& =& \frac{1}{T_1}\\ T_2& =& 1.5T_1\end{array}$

Hence, the temperature of the gas increases 1.5Times.