A tank is filled with an ideal gas at 400 and pressure of 1.00 . The…
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Question “A tank is filled with an ideal gas at 400 and pressure of 1.00 . The…”
thegas in the tank doubles. What is the temperature of the
gas?
Answer
Ideal gas law is the solution to this problem.
To calculate the final temperature, first use the initial temperature, the initial pressure, or the final pressure.
To calculate the temperature of the gas, you can use the ideal gas law equation.
The ideal gas law can be described as:
PV = nRTHere, P is pressure, V the volume, n the number of moles and R the gas constant are all used. T is temperature.
The expression is:
\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}Here, {P_1}
To get the final gas temperature, you can modify the equation above.
{T_2} = \frac{{{P_2}{T_1}}}{{{P_1}}}The expression for the number moles found in the container prior to gas leakage is
n = {n_1}Here, $math_tag_5 is the number of moles.
The expression for the number moles found in the container following a leakage of gas is
n = \frac{{{n_1}}}{2}The ideal gas law that prevents gas from leaking is described as,
{P_2}V = nR{T_2}Substitute $math_tag_8
{P_2}V = {n_1}R{T_2}The ideal gas law is one that prevents the release of gas.
{P_3}V = nR{T_3}Here, {P_3}
Substitute \[n_1 \mathord\left/\\ \vphantom n_1 2 \right.\\ \kern-\nulldelimiterspace 2\] for n.
{P_3}V = \left( {\frac{{{n_1}}}{2}} \right)R{T_3}
(1)
The final temperature of the gas can be expressed as:
{T_2} = \frac{{{P_2}{T_1}}}{{{P_1}}}Substitute $math_tag_14
\begin{array}{c}\\{T_2} = \frac{{\left( {2.00\,{\rm{atm}}} \right)\left( {400\,{\rm{K}}} \right)}}{{\left( {1.00\,{\rm{atm}}} \right)}}\\\\ = 800\,{\rm{K}}\\\end{array}Therefore, $math_tag_16 is the best option
(2)
The ideal gas law, before any gas leakage, is described as
{P_2}V = {n_1}R{T_2}Arrange the above according to its constant.
\frac{V}{{R{n_1}}} = \frac{{{T_2}}}{{{P_2}}}After a leakage, the ideal gas law is:
{P_3}V = \left( {\frac{{{n_1}}}{2}} \right)R{T_3}Arrange the above according to its constant.
\frac{V}{{R{n_1}}} = \frac{{{T_3}}}{{2{P_3}}}Add the equations (4) and 5.
\frac{{{T_2}}}{{{P_2}}} = \frac{{{T_3}}}{{2{P_3}}}To find the temperature of the gas following leakage, you can modify the equation.
{T_3} = \frac{{2{P_3}{T_2}}}{{{P_2}}}Substitute $math_tag_23
$math_tag_24 Ans: Part 1
The gas’s final temperature is $math_tag_25 Part 2
After a leakage, the temperature of the gas is $math_tag_25
Conclusion
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