Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • Three moles of an ideal gas being initially at a temperature <img alt="\mathrm{T}_0=273 \mathrm{~K}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BT%7D_0%3D273%20%5Cmathrm%7B%7EK

Three moles of an ideal gas being initially at a temperature \mathrm{T}_0=273 \mathrm{~K} were isothermally expanded \mathrm{\eta=5.0} time its initial volume and then isochorically heated so that the pressure in the final state became equal to that in the initial state. The total amount of heat transferred to the gas during the process equals \mathrm{Q}=80 \mathrm{KJ}. Find the ratio \mathrm{\gamma=\mathrm{Cp} / \mathrm{Cv}} for this gas.
 

Option: 1

1.23


Option: 2

1.33


Option: 3

1.4


Option: 4

1.5


Answers (1)

best_answer

In isothermal process, the heat transferrd to the gas is given by

\mathrm{ \mathbf{Q}_{\mathbf{l}}=\mathbf{n R T} \mathbf{R} \ln \left(\mathbf{V}_2 / \mathbf{V}_{\mathbf{l}}\right)=\mathbf{n R T} \mathbf{T}_0 \ln \eta }          (1)

\mathrm{ {\left[\therefore \quad \eta=\left(\mathbf{V}_2 / \mathbf{V}_1\right)=\left(\mathrm{P}_1 / \mathbf{P}_2\right)\right. \text { ] }} }

\mathrm{ \text { In isochroric process, } \mathrm{Q}_2=\Delta \mathrm{U} \quad(\mathrm{W}=\mathbf{0}) }

\mathrm{ \therefore \quad \mathrm{Q}_2=\mathrm{nCv} \Delta \mathrm{T}=\mathbf{n}\{\mathrm{R} /(\gamma-\mathbf{1})\} \Delta \mathrm{T} }          (2)

\mathrm{ \text { Now } \frac{P_2}{P_1}=\frac{T_0}{T} \quad \text { or } \quad T=T_0\left(\frac{P_1}{P_2}\right)=\eta T_0 }

\mathrm{ \therefore \quad \Delta T=\eta T_0-T_0=(\eta-1) T_0 }           (3)

substituting the value of \mathrm{\Delta T} from equation (3) in equation (2), we get

\mathrm{ Q_2=n\left(\frac{R}{\gamma-1}\right)(\eta-1) T_0 }

\mathrm{ \therefore \quad Q=n R T_0 \ln \eta+n\left(\frac{R}{\gamma-1}\right)(\eta-1) T_0 }

\mathrm{ \text { or } \frac{Q}{\mathrm{nRT}_0}-\ln \eta=\left(\frac{\eta-1}{\gamma-1}\right) }

\mathrm{ \text { or } \quad \gamma-1=\frac{\eta-1}{\frac{Q}{\mathrm{nRT}}-\ln \eta} }

\mathrm{ \therefore \quad \gamma=1+\frac{\eta-1}{\frac{Q}{\mathrm{nRT_{0 }}}-\ln \eta} }

Substituting given values, we get

\mathrm{ \gamma=1+\frac{(5-1)}{\frac{80 \times 10^3}{3 \times 8.3 \times 273}-\ln 5} }

Solving we get \mathrm{ \gamma=1.4 }

Posted by

manish

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

NTA warns NEET 2026 aspirants against ‘paper leak’ claims circulating on social media
anam-khan

NEET UG 2026 FAQs: NTA answers queries on ID proof, dress code, OMR rules, exam guidelines
anam-khan

NEET UG 2026 Cut-off: Arunachal Pradesh MBBS opening and closing ranks, past-year trends

Latest Question