Correct answer is (B)

For all process,

E_AB = E_AC = E_AD

W_AB > 0 as volume increases.

W_AC = 0 as volume constant.

W_AD < 0 as volume decreases.

Correct answer is (A)

At constant pressure,

W = P$\mathrm{\Delta }$V = nR$\mathrm{\Delta }$T

At constant pressure, heat supplied

Q = nC_p$\mathrm{\Delta }$T

$\frac{W}{Q}=\frac{R}{C_p}$

For diatomic gas, C_p = $\frac{7}{2}R$

$\therefore$ $\frac{10}{Q}=\frac{R}{\frac{7}{2}R}$

$\Rightarrow$ Q = 10 $\times$ $\frac{7}{2}$ = 35 J

Correct answer is (B)

For the process (c – a), $\mathrm{\Delta }$U_ca = – 180 J

For process (b – c) $\to$ Isochoric (W_bc = 0)

$\therefore \mathrm{\Delta }$U = 60 J

Heat absorbs along (a – b), Q_ab = 250 J

Also $\therefore \mathrm{\Delta }$ U_cycle = 0

$\therefore \mathrm{\Delta }$ U_ab = 120 J

So W_{a $\to$ b} = 130 J

Total work done from (a $\to$ b $\to$ c)

= W_ab + W_bc = 130 J

Correct answer is (A)

Heat supplied at constant volume
Q = nC_V$\mathrm{\Delta }$T

and heat supplied at constant pressure
Q' = nC_P$\mathrm{\Delta }$T

$Q^{\prime }=\frac{C_P}{C_V}Q=\left(1+\frac{2}{5}\right)Q=\frac{7}{5}Q$

Correct answer is (C)

$\mathrm{\Delta }Q=n{C}_v$    $\mathrm{\Delta }T=n\frac{3}{2}R\mathrm{\Delta }T$

$=\left(\frac{67.2}{22.4}\right)\left(\frac{3}{2}\times 8.31\right)\left(20\right)$

$\approx 748J$