Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. If f is the number of degre

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. If f is the number of degree of freedom of gas molecules and W is the work done by the gas during the expansion, then

Option: 1

f = 3


Option: 2

f = 5


Option: 3

\mathrm{W=\frac{3 P V}{4}}


Option: 4

\mathrm{W=\frac{3 P V}{2}}


Answers (1)

best_answer

For an adiabatic change the relation between T and V is

\mathrm{T V^{(\gamma-1)}=\text { constant } ; \gamma=\frac{C_p}{C_v}}

\mathrm{T V^{(\gamma-1)}=T^{\prime} V^{\prime(\gamma-1)} \text { or }\left(\frac{V^{\prime}}{V}\right)^{(\gamma-1)}=\frac{T}{T^{\prime}}}

\mathrm{\text { Given } V^{\prime}=5.66 V \text { and } T^{\prime}=\frac{T}{2} \text {. Therefore, }}

\mathrm{(5.66)^{(\gamma-1)}=2}

Taking logarithm of both sides, we have

\mathrm{\begin{aligned} & (\gamma-1) \log (5.66)=\log (2) \\ & \text { Or } \gamma=1+\frac{\log (2)}{\log (5.66)}=1+\frac{0.3010}{0.7528} \\ & =1+0.4=1.4 \end{aligned}}

Since γ = 1.4, the gas is diatomic. For a diatomic gas, the number of degrees of freedom
of the molecules = 5
We know that the work done by the gas during adiabatic expansion is given by

\mathrm{W=\frac{1}{(\gamma-1)}\left(P V-P^{\prime} V^{\prime}\right)}          ....[1]

Where pressure\mathrm{P^{\prime}}′ after expansion I obtained from the relation

\mathrm{\begin{aligned} & \frac{P^{\prime} V^{\prime}}{T^{\prime}}=\frac{P V}{T} \\ & \text { Or } P^{\prime}=P \times \frac{V}{V^{\prime}} \times \frac{T^{\prime}}{T} \\ & =P \times \frac{V}{5.66 V} \times \frac{T / 2}{T}=\frac{T / 2}{T}=\frac{P}{11.32} \end{aligned}}

\mathrm{\text { Putting } \gamma=1.4, V^{\prime}=5.66 \mathrm{~V} \text { and } P^{\prime}=\frac{P}{11.32} \text { in }}

In Eq. (1), we have

\mathrm{\begin{aligned} & W=\frac{1}{(1.4-1)}\left(P V-\frac{P}{11.32} \times 5.66 V\right) \\ & =\frac{1}{0.4}\left(P V-\frac{1}{2} P V\right)=1.25 P V \end{aligned}}

So the correct choices are (b) and (c)

 

Posted by

qnaprep

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG 2025 'safe': All state governments on alert, NTA cautions against paper leak, misleading rumours
anam-khan

NEET-UG mock drills underway across centres ahead of May 4 entrance
anam-khan

NEET probe: NMC cancels admission of 14 students, orders suspension of 26 more

Latest Question