Question

One mole of a certain ideal gas obtains an amount of heat $Q \, = \, 1.60 \, kJ$ when its temperature is increased by $\Delta T \, = \, 72 \, K$ , keeping its pressure constant. The value of $\frac{C_{P}}{C_{V}}$ for the gas is

• A. $1.60$
• B. $1.40$
• C. $1.50$
• D. $1.30$

Answer 1

Answer: (A)

By the first law of thermodynamics
$ \Delta Q =\Delta U +\Delta W $
In an isobaric process
$ \Delta Q = C _{ p } \Delta T $
and $\Delta U = C _{ v } \Delta T $ (always)
$ \therefore C _{ p } \Delta T = C _{ v } \Delta T +\Delta W $
or $\Delta W =\left( C _{ p }- C _{ v }\right) \Delta T$
or $\Delta W = R \Delta T \left(\because C _{ p }- C _{ v }= R \right)$
$ \begin{aligned} \therefore & \Delta W =8.3 \times 72=597.6 J \\ & \Delta Q = C _{ p } \Delta T \\ \therefore & 1.6 \times 1000= C _{ p } \times 72 \\ \Rightarrow & C _{ p }=\frac{1.6 \times 1000}{72}=22.2 J mol ^{-1} K ^{-1} \end{aligned} $
$ \Delta U =\Delta Q =\Delta W =1.6 \times 1000-597.6 J =1002.4 J $
But $\Delta U = C _{ v } \Delta T$
$ \therefore C _{ v }=\frac{1002.4}{72}=13.9 J mol ^{-1} K ^{-1} $
$∴ \, \, \, \gamma = \frac{\text{C}_{\text{p}}}{\text{C}_{\text{v}}} = \frac{\text{22.2}}{\text{13.9}} = \text{1.60}$