Question

If Q amount of heat is given to a diatomic ideal gas in a process in which the gas perform a work (2Q)/(3) on its surrounding. Find the molar heat capacity (in terms of R) for the process.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Molar\:heat\:capacity=7.5R}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies Heat \: supplied (Q) = Q \\ \\ \tt: \implies Work \: done(W) = \frac{2Q}{3} \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt: \implies Molar \: heat \: capacity( C_{vm}) = ?$

• According to given question :

$\bold{From \: first \: law \: of \: thermodynamics} \\ \tt: \implies Q = \Delta U + W \\ \\ \tt: \implies Q = \Delta U + \frac{2Q }{3} \\ \\ \tt: \implies \Delta U= Q- \frac{2Q}{3} \\ \\ \tt: \implies \Delta U = \frac{Q}{3} - - - - - (1) \\ \\ \bold{For \: diatomic \: ideal \: gas} \\ \tt: \implies \Delta U = nr\Delta T \\ \\ \tt: \implies \frac{Q}{3} = n \times \frac{5}{2}R \Delta T \\ \\ \tt : \implies Q =7.5\: nR\Delta T - - - - - (2)\\ \\ \bold{For \: molar \: heat \: capacity} \\ \tt: \implies Q = n C_{vm}\Delta T \\ \\ \tt: \implies 7.5 \: nR\Delta T = n C_{vm}\Delta T \\ \\ \green{\tt: \implies C_{vm} = 7.5 \: R}$

Answer 2

Answer:

$\green{ \tt{molar \: heat \: capacity( c_{vm} ) = 7.5 \: R}}$

Explanation:

$\tt \to q = \Delta U + W \: \: \: (from \: \: F.L.O.T)\\ \\ \tt \to Q = \Delta U+ \frac{2Q}{3} \\ \\ \tt \to Q- \frac{2Q}{3} = \Delta U \\ \\ \tt \to \Delta U= \frac{Q}{3} - - - - - (1) \\ \\ \bold{For \: diatomic \: gas} \\ \tt \to \Delta U = \frac{5}{2}nR \Delta T \\ \\ \tt \to \frac{Q}{3} = \frac{5}{2} nR\Delta T \\ \\ \tt \to Q = \frac{15}{2} nR \Delta T - - - - - (2)\\ \\ \bold{As \: according \: to \: molar \: heat \: capacity \: formula} \\ \tt \to Q= n c_{vm} \Delta T - - - - - (3) \\ \\ \bold{From \: (2) \: and \: (3)} \\ \tt \to \frac{15}{2} nR\Delta T= n C_{vm} \Delta T\\ \\ \green{\tt \to C_{vm} = \frac{15}{2} R}$