JEE MAIN - Chemistry (2019 - 8th April Evening Slot - No. 8)
5 moles of an ideal gas at 100 K are allowed
to undergo reversible compression till its
temperature becomes 200 K.
If CV = 28 JK–1mol–1, calculate $$\Delta $$U and $$\Delta $$pV for
this process. (R = 8.0 JK–1 mol–1]
$$\Delta $$U = 14 kJ; $$\Delta $$(pV) = 4 kJ
$$\Delta $$U = 2.8 kJ; $$\Delta $$(pV) = 0.8 kJ
$$\Delta $$U = 14 kJ; $$\Delta $$(pV) = 18 kJ
$$\Delta $$U = 14 kJ; $$\Delta $$(pV) = 0.8 kJ
Explanation
For ideal gas,
$$\Delta $$U = nCv$$\Delta $$T
= 5 $$ \times $$ 28 $$ \times $$ (200 - 100)
= 14 kJ
We know,
PV = nRT
$$ \therefore $$ $$\Delta $$(PV) = nR$$\Delta $$T = 5 $$ \times $$ 8 $$ \times $$ 100 = 4 kJ
$$\Delta $$U = nCv$$\Delta $$T
= 5 $$ \times $$ 28 $$ \times $$ (200 - 100)
= 14 kJ
We know,
PV = nRT
$$ \therefore $$ $$\Delta $$(PV) = nR$$\Delta $$T = 5 $$ \times $$ 8 $$ \times $$ 100 = 4 kJ
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