A monatomic gas initially at 18∘C is compressed adiabatically to 18 of its original volume. The temperature after compression will be
A
180∘C
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B
144∘C
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C
891∘C
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D
887∘C
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Solution
The correct option is C891∘C For an adiabatic process, T1V1γ−1=T2V2γ−1 For monoatomic gas, γ=53 Given, T1=18+273=291K, V2=V18 291×V53−11=T2(V18)53−1 T2=291×82/3 T2=291×4 T2=1164K=891∘C