An ideal monatomic gas is contained in a cylinder with a movable piston so that the gas can do work on the outside worl…
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Question “An ideal monatomic gas is contained in a cylinder with a movable piston so that the gas can do work on the outside worl…”
An ideal monatomic gas is contained in a cylinder with a movable
piston so that the gas can do work on the outside world, and heat
can be added or removed as necessary. The figure shows various
paths that the gas might take in expanding from an initial state
whose pressure, volume, and temperature are , , and respectively. The gas expands to a state with
final volume . For some answers it will be convenient to
generalize your results by using the variable , which
is the ratio of final to initial volumes (equal to 4 for the
expansions shown in the figure.)
The figure shows several possible paths of the system in the
pV plane. Although there are an infinite number of paths
possible, several of those shown are special because one of their
state variables remains constant during the expansion. These have
the following names:
- Adiabiatic: No heat is added or removed during the
expansion. - Isobaric: The pressure remains constant during the
expansion. - Isothermal: The temperature remains constant during
the expansion.
during an isobaric expansion?
|
an isothermal process?
|
pV plot ____________.
|
adiabatic expansion?
|
Answer
To solve this problem, you can use the isobaric, adiabatic and isothermal concepts.
Determine the temperature dependence of the isobaric processes by analysing the formula.
Analyze the isothermal condition and determine which processes are itothermal.
To determine which area of the work is being done, analyze the expressions.
An isothermal process is one where the temperature of the gas is maintained constant throughout the process. Here is the condition for an isothermal process:
pV = KHere, p
An isobaric process refers to one where pressure does not change throughout the entire process. This is the condition of an isobaric procedure.
\frac{V}{T} = KHere, V
A adiabatic process is one where there is no heat transfer. An adiabatic process can be described as follows:
p{V^\gamma } = KHere,\gamma
Any thermodynamic process requires work that is:
W = \int_{{V_i}}^{{V_f}} {pdV}Here, W
(A)
You can use this method to achieve isobaric effects
\frac{V}{T} = KThis means that the temperature of the gas directly affects its volume. Temperature of the gas rises with volume (volume increases while expanding) for an isobaric expansion.
(C)
The following is the condition for isothermal extension:
pV = KYou can write it as follows.
{p_1}{V_1} = {p_2}{V_2}This condition suggests that the graph p is not correct.
Find out which curve meets the conditions for isothermal processes.
$math_tag_12 is the right place to go for paths B, C and D.
Now, calculate $math_tag_13
For the B path:
\begin{array}{c}\\{P_2}{V_2} = \left( {0.625} \right)\left( 4 \right)\\\\ = 2.5\\\end{array}So, {P_1}{V_1} \ne {P_2}{V_2}
Path Bis is therefore not an isothermal process.
You can follow the D path
\begin{array}{c}\\{P_2}{V_2} = \left( {0.140} \right)\left( 4 \right)\\\\ = 0.56\\\end{array}Path D is not isothermal.
For the path C:
\begin{array}{c}\\{P_2}{V_2} = \left( {0.25} \right)\left( 4 \right)\\\\ = 1\\\end{array}So, {P_1}{V_1} = {P_2}{V_2}
Path C therefore is an isothermal process.
[Part C]
Part B
(D)
Here is the work that was done for any thermodynamic process:
W = \int_{{V_i}}^{{V_f}} {pdV}The area under the curve for $math_tag_20 is the only thing in the equation.
(E)
Here are the results of work done for an isobaric procedure.
W = p\left( {{V_2} - {V_1}} \right)Substitute, {p_0} \begin{array}{c}\\W = {p_0}\left( {{R_v}{V_0} - {V_0}} \right)\\\\ = {p_0}{V_0}\left( {{R_v} - 1} \right)\\\end{array}
[Part E]
(G)
The following conditions are required for adiabatic processes:
{p_1}V_1^\gamma = {p_2}V_2^\gammaMonoatomic gas: $math_tag_25
{p_1}V_1^{1.4} = {p_2}V_2^{1.4}All processes
\begin{array}{c}\\{p_1}V_1^{1.4} = \left( 1 \right){\left( 1 \right)^{1.4}}\\\\ = 1\\\end{array}Now, calculate $math_tag_28
To go to A:
\begin{array}{c}\\{P_2}V_2^{1.4} = \left( {1.00} \right){\left( 4 \right)^{1.4}}\\\\ = 6.96\\\end{array}For the B path:
\begin{array}{c}\\{P_2}V_2^{1.4} = \left( {0.625} \right){\left( 4 \right)^{1.4}}\\\\ = 4.35\\\end{array}For the path C:
\begin{array}{c}\\{P_2}V_2^{1.4} = \left( {0.25} \right){\left( 4 \right)^{1.4}}\\\\ = 1.74\\\end{array}You can follow the D path
\begin{array}{c}\\{P_2}V_2^{1.4} = \left( {0.140} \right){\left( 4 \right)^{1.4}}\\\\ = 0.975\\\end{array}Therefore, path D is the only way to adiabatic expansion.
Part A – Ans
An isobaric expansion causes the gas to heat up.
Conclusion
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