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

Solution

## The correct option is C $$a \ ln(\eta) -\left (\frac {\eta -1}{\gamma-1}\right )RT_0$$Q=w+Auw=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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More