When one mole of an ideal gas is compressed to half of the initial volume and simultaneously heated to twice of its initial temperature, the change in enthalpy $(Δ S)$ is (Considering the process to be isobaric)

C_{v} l n 2

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C_{p} l n 2

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R l n 2

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(C_{v} - R) l n 2

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Solution

The correct option is D $(C_{v} - R) l n 2$
By using the formula,
$Δ S = C_{v} l n (\frac{T_{2}}{T_{1}}) + R l n \frac{V_{2}}{V_{1}} As the gas is heated to twice its temperature \Rightarrow T_{2} = 2 T_{1} And is compressed to half of its initial volume V_{2} = \frac{V_{1}}{2}$
Substituting the values we get,
$Δ S = C_{v} l n 2 + R \times l n \frac{1}{2} = (C_{v} - R) l n 2$

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When one mole of an ideal gas is compressed to half of the initial volume and simultaneously heated to twice of its initial temperature, the change in enthalpy (ΔS) is (Considering the process to be isobaric)

The correct option is D (Cv−R) ln 2By using the formula, ΔS=Cvln(T2T1)+R lnV2V1As the gas is heated to twice its temperature⇒T2=2T1And is compressed to half of its initial volume V2=V12 Substituting the values we get, ΔS=Cv ln 2+R×ln12=(Cv−R) ln 2

When one mole of an ideal gas is compressed to half of the initial volume and simultaneously heated to twice of its initial temperature, the change in enthalpy $(Δ S)$ is (Considering the process to be isobaric)