Question

One mole of a monoatomic ideal gas undergoes the process A® B as in the given P-V diagram. The specific heat for this process is :

• A. 3R/2
• B. 15R/7
• C. 30R/7
• D. 20R/7

Answer 1

Answer: (B)

15R/7

Answer 2

DQ = DU + W (from 1^st law)

Here DU = nC_VDT

Temperature at A,

T_A =$\frac{2P_{0}V_{0}}{R}$ & at B, T_B =$\frac{16P_{0}V_{0}}{R}$

So DU =$\frac{3}{2}$R$\left\lbrack \frac{14P_{0}V_{0}}{R} \right\rbrack$= 21 P₀V₀

W = (2P₀) (3V₀) +$\frac{1}{2}$ (3V₀) (2P₀) ̃ 9P₀V₀

So DQ = 21 P₀V₀ + 9P₀V₀

̃ 30 P₀V₀

.DQ = nCD

C =$\frac{\Delta Q}{n\Delta T}$=$\frac{30P_{0}V_{0}}{14P_{0}V_{0}/R}$=$\frac{15}{7}$R