Find the change in the internal energy when 15 -160-gm of air is heated from - 0-oC- to - 5-oC- The specific -160-heat of air at constant volume is 0-2 cal-gm- -oC-75 Cal30 Cal15 Cal105 Cal

The correct option is C 15 Cal #160;heat = Δ Q =Mass #160; × #160;specific heat × temperature change= 15 × 0.2 × 5 cal =15 calAlso, is ...

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Standard XII

Chemistry

Work Done in Isothermal Reversible Process

Find the chan...

Question

Find the change in the internal energy when 15 gm of air is heated from $0^{o} C$ to $5^{o} C$ . The specific heat of air at constant volume is 0.2 cal/gm $^{o} C$ .

75 Cal

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30 Cal

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15 Cal

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105 Cal

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Solution

The correct option is C 15 Cal
heat = $Δ Q$ =Mass $\times$ specific heat $\times$ temperature change= $15 \times 0.2 \times 5 c a l$ =15 cal
Also, isochoric process implies work done =0
Applying first law, Change in internal energy= $Δ Q$ =15 cal

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