Question

A gas at temperature $T_0$and atmospheric pressure fills a cylinder. The gas is transferred to a new cylinder with three times the volume, after which the pressure is half the original pressure. What is the new temperature of the gas?

Short answer

The new temperature of the gas is$\frac{\mathbf{3}}{\mathbf{2}}{\mathit{T}}_{\mathbf{0}}$

Step-by-step solution

: Given information and Formula used 

Given : Gas temperature :$T_0$

Pressure : Atmospheric pressure

New cylinder volume : Three times the original volume

New pressure : Half the original pressure

Theory used :

Combined gas lawstates that $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

That is equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Standard temperature and pressure, STP, is $273K\mathrm{and}1$atmosphere.

Finding the new temperature of the gas .

Since the number of moles did not change in this process, we can write that.
$\frac{p_0{V}_0}{T_0}=\frac{p_1{V}_1}{T_1}$.
Since we know that $V₁=3V_0\mathrm{and}p₁=p_0/2$, we can substitute and get
role="math" localid="1648407236426" $$\begin{gathered} \frac{p_0{V}_0}{T_0}=\frac{p_{0}3V_0}{2T_1} \\ \Rightarrow {\mathit{T}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}{\mathit{T}}_{\mathbf{0}} \end{gathered}$$