Starting at temperature $$300$$ K , one mole of an ideal , diatomic gas γ$$=1.4$$ is first compressed adiabatically from volume $$V1$$ to $$V2=V116$$ . It is then allowed to expand isobarically to volume $$2V2$$. If all the processes are the $$quasi-static$$ then the final temperature of the gas (in$$ K $$) is $$(to$$ the nearest $$integer)___$$

Question

Starting at temperature $$300$$ K , one mole of an ideal , diatomic gas γ$$=1.4$$ is first compressed adiabatically from volume $$V1$$ to $$V2=V116$$ . It is then allowed to expand isobarically to volume $$2V2$$. If all the processes are the $$quasi-static$$ then the final temperature of the gas (in$$ K $$) is $$(to$$ the nearest $$integer)___$$

Infinity Learn · Accepted Answer

During adiabatic processTVγ−$$1=constant$$    ⇒$$T1V1$$γ−$$1=T2$$.$$V2$$γ−$$1$$    ⇒$$T2=T1(V1V2)$$γ−$$1=300(16)75$$−$$1=T2=300(24)2/5$$  $$=T2=300$$×$$24$$×$$25During$$ isobaric $$processV=nRTP$$     ⇒V∝T     ⇒$$V2V3=T2T3$$ ⇒$$V22V2=300$$×$$24$$×$$25T3$$     ⇒$$T3=2$$×$$300$$×$$28/5$$     $$=1818.85$$  $$=1819$$ K

Thermodynamic processes

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Thermodynamic processes

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During adiabatic processTVγ−1=constant ⇒T1V1γ−1=T2.V2γ−1 ⇒T2=T1(V1V2)γ−1=300(16)75−1=T2=300(24)2/5 =T2=300×24×25During isobaric processV=nRTP ⇒V∝T ⇒V2V3=T2T3 ⇒V22V2=300×24×25T3 ⇒T3=2×300×28/5 =1818.85 =1819 K

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