# NEET Physics Thermodynamics Questions Solved The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its volume at 27°C, if the value of γ = 1.4, then the new temperature will be -

(1) 350 × 40.4 K

(2) 300 × 40.4 K

(3) 150 × 40.4 K

(4) None of these

Concept Videos :-

#3 | Thermodynamic Processes
#4 | Isobaric Process
#5 | Isochoric & Isothermal Process
#6 | Interconversion of Graphs
#7 | Solved Example: 1
#8 | Solved Example: 2
#9 | Adiabatic Process
#10 | Solved Example: 3
#11 | Graphs in Adiabatic Process: 1
#12 | Graphs in Adiabatic Process: 2

Concept Questions :-

Types of processes

(2) For adiabatic change $T{V}^{\gamma -1}$ = constant

$\frac{{T}_{2}}{{T}_{1}}={\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma -1}$${T}_{2}={\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma -1}×{T}_{1}$

${T}_{2}={\left(\frac{V}{V/4}\right)}^{1.4-1}×300=300×{\left(4\right)}^{0.4}K$

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