Heat and Thermodynamics 5 Question 52

53. At 27C two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V.

Calculate

(1996, 5M)

(a) the final temperature of the gas,

(b) change in its internal energy,

(c) the work done by the gas during this process.

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Answer:

Correct Answer: 53. (a) 189K (b) 2767J (c) 2767J

Solution:

  1. Given, T1=27C=300K,V1=V,V2=2V

(a) Final temperature

In adiabatic process, TVγ1= constant

T1V1γ1=T2V2γ1

or T2=T1(V1V2)γ1=300(V2V)5/31

γ=53 for a monoatomic gas T2189K

(b) Change in internal energy,

ΔU=nCVΔTΔU=(2)(32R)(T2T1)ΔU=2(32)(8.31)(189300)JΔU=2767J

(c) Work done

Process is adiabatic, therefore ΔQ=0

and from first law of thermodynamics,

ΔQ=ΔW+ΔUΔW=ΔU=(2767J)ΔW=2767J



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