Question

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[Chemistry]

It is found, on heating a gas, its volume increases by 50% and pressure decreases to 60% of its original value. If the original temperature was -15°C, find the temperature to which it was heated ?​

Answer 1

Answer:

(232.2- 273) $\\\tt ^{\circ}C$ = -40.8 $\\\tt ^{\circ}C$

Formulas/Laws used:

• Ideal gas law: PV= nRT

Where P = pressure,  V= volumes = n = number of moles, R= Gas constant and T = temperature

and all the units are of their respective SI's.

Logic:

Assuming that the number of moles are constant.

• Ideal gas law at initial temperature

• Ideal gas law relation at final temperature

• Divide them to eliminate n and R.

Note: Make sure Units are of SI.

Explanation:

Initial conditions:

Let,

initital pressure( $\\\tt P_{i}$ ) be  $\\\tt P_{0}$

initial volume($\\\tt V_{i}$) be $\\\tt V_{0}$  

initla temperature ($\\\tt T_{i}$) be $\\\tt T_{0}$

For these two conditions Ideal gas law relation,

$\\\tt P_{i} V_{i} = N R T_{i } \\ \Rightarrow P_{0} V_{0} = N R (273-15) \\ \Rightarrow P_{0} V_{0} = N R (258)$

Final conditions:

Pressure ($\\\tt P_{f}$) = $\\\tt P_{0} + \dfrac{1}{2} P_{0}= \dfrac{3}{2} P_{0}$

Volume( $\\\tt V_{f}$)=$\\\tt \dfrac{60}{100} V_{0} = \dfrac{3}{5} V_{0}$

Final temperature = $\\\tt T_{f}$

Using Ideal gas law for these conditions,

$\\\tt P_{f} V_{f} = N R T_{f}\\ \\\Rightarrow \dfrac{9 P_{0} V_{0}}{10} = N RT_{f}$

#eleminating n and R

Dividing initial condition by final

$\\\tt \dfrac{P_{i} V_{i}}{P_{f} V_{f}} = \dfrac{N R T_{i}}{N R T_{f}} \\ \Rightarrow \dfrac{P_{0} V_{0}}{\dfrac{9 P_{0} V_{0}}{10}} = \dfrac{N R (258)}{N R T_{f}} \\ \Rightarrow \dfrac{10}{9 } = \dfrac{(258)}{T_{f}} \\ \Rightarrow T_{f} = \dfrac{9}{10 } (258) = 232.2 K$

Answer 2

Explanation:

Answer:

$(232.2- 273) \begin{gathered}\\\tt ^{\circ}C\end{gathered}$

$= -40.8 \begin{gathered}\\\tt ^{\circ}C\end{gathered}$

Formulas/Laws used:

Ideal gas law: PV= nRT

Where P = pressure, V= volumes = n = number of moles, R= Gas constant and T = temperature

and all the units are of their respective SI's.

Logic:

Assuming that the number of moles are constant.

Ideal gas law at initial temperature

Ideal gas law relation at final temperature

Divide them to eliminate n and R.

Note: Make sure Units are of SI.

Explanation:

Initial conditions:

Let,

initital pressure

$( \begin{gathered}\\\tt P_{i}\end{gathered}$

$\begin{gathered}\\\tt P_{0}\end{gathered}$

initial volume

$(\begin{gathered}\\\tt V_{i}\end{gathered}$

$\begin{gathered}\\\tt V_{0}\end{gathered}$

$initla temperature (\begin{gathered}\\\tt T_{i}\end{gathered}$

$\begin{gathered}\\\tt T_{0}\end{gathered}$

For these two conditions Ideal gas law relation,

$\begin{gathered}\\\tt P_{i} V_{i} = N R T_{i } \\ \Rightarrow P_{0} V_{0} = N R (273-15) \\ \Rightarrow P_{0} V_{0} = N R (258)\end{gathered}$

Final conditions:

$Pressure (\begin{gathered}\\\tt P_{f}\end{gathered}$

$) = \begin{gathered}\\\tt P_{0} + \dfrac{1}{2} P_{0}= \dfrac{3}{2} P_{0}\end{gathered}$

$Volume( \begin{gathered}\\\tt V_{f}\end{gathered}$

$)=\begin{gathered}\\\tt \dfrac{60}{100} V_{0} = \dfrac{3}{5} V_{0}\end{gathered}$

Final temperature = $\begin{gathered}\\\tt T_{f}\end{gathered}$

Using Ideal gas law for these conditions,

$\begin{gathered}\\\tt P_{f} V_{f} = N R T_{f}\\ \\\Rightarrow \dfrac{9 P_{0} V_{0}}{10} = N RT_{f}\end{gathered}$

#eleminating n and R

Dividing initial condition by final

$\begin{gathered}\\\tt \dfrac{P_{i} V_{i}}{P_{f} V_{f}} = \dfrac{N R T_{i}}{N R T_{f}} \\ \Rightarrow \dfrac{P_{0} V_{0}}{\dfrac{9 P_{0} V_{0}}{10}} = \dfrac{N R (258)}{N R T_{f}} \\ \Rightarrow \dfrac{10}{9 } = \dfrac{(258)}{T_{f}} \\ \Rightarrow T_{f} = \dfrac{9}{10 } (258) = 232.2 K\end{gathered}$