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Question

The molar heat capacity C for an ideal gas going through a process is given by $$C=\dfrac {a}{T}$$, where 'a' is a constant. If $$\gamma=\dfrac {C_P}{C_V}$$, the work done by one mole of gas during heating from $$T_0$$ and $$\eta T_0$$ will be:


A
a ln(η)
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B
1a ln(η)
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C
a ln(η)(η1γ1)RT0
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D
a ln(η)(γ1)RT0
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Solution

The correct option is C $$a \ ln(\eta) -\left (\frac {\eta -1}{\gamma-1}\right )RT_0$$
Q=w+Au
w=Q-Au                      Q=cdt  (for 1 mole)
Q=$$\int^{T1}_{T0}[\dfrac {a}{t}dT]=a\ln n$$
$$Au=\eta\dfrac{R}{\gamma-1}At $$    (here n=1mole)
$$Au=\dfrac{R}{\gamma-1}T_0(\gamma-1)$$
$$w=a\ln(\gamma)-(\dfrac{\eta-1}{\gamma-1})RT_0$$

Physics

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