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Question

Physics Question on Thermodynamics

During an adiabatic compression, 830J830\, J of work is done on 22 moles of a diatomic ideal gas to reduce its volume by 50%50\%. The change in its temperature is nearly : (R=8.3JK1mol1)(R = 8.3 JK^{-1}\, mol^{-1})

A

40 K

B

33 K

C

20 K

D

14 K

Answer

20 K

Explanation

Solution

The work done in adiabatic process is given as
W=nRΔTγ1 W=\frac{n R \Delta T}{\gamma-1}
or
830=2×8.3×ΔT1.41830=\frac{2 \times 8.3 \times \Delta T}{1.4-1}
ΔT=20K \Delta T=20 K