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NEET PHYSICSMedium

The volume (VV) of a monatomic gas varies with its temperature (TT), as shown in the graph. The ratio of work done by the gas to the heat absorbed by it when it undergoes a change from state AA to state BB will be:

A

25\frac{2}{5}

B

23\frac{2}{3}

C

13\frac{1}{3}

D

27\frac{2}{7}

Step-by-Step Solution

From the description, the graph of Volume (VV) versus Temperature (TT) is a straight line passing through the origin (implied by the variation VTV \propto T). This indicates that the process is isobaric (constant pressure).

  1. Work Done (WW): In an isobaric process, the work done by the gas is given by: W=PΔV=nRΔTW = P\Delta V = nR\Delta T

  2. Heat Absorbed (QQ): The heat absorbed at constant pressure is given by: Q=nCpΔTQ = nC_p\Delta T Where CpC_p is the molar heat capacity at constant pressure.

  3. Specific Heat for Monatomic Gas: For a monatomic ideal gas, the molar heat capacity at constant pressure is: Cp=52RC_p = \frac{5}{2}R (Source: Appendix, Molar Heat Capacities )

  4. Calculate Ratio: Ratio=Work DoneHeat Absorbed=WQ=nRΔTnCpΔT=RCp\text{Ratio} = \frac{\text{Work Done}}{\text{Heat Absorbed}} = \frac{W}{Q} = \frac{nR\Delta T}{nC_p\Delta T} = \frac{R}{C_p} Substituting Cp=52RC_p = \frac{5}{2}R: Ratio=R52R=25\text{Ratio} = \frac{R}{\frac{5}{2}R} = \frac{2}{5}

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