since gas undergoes isobaric process
$\Rightarrow \Delta \mathrm{Q}=\mathrm{n} \frac{7}{2} \mathrm{R\Delta\,T}=\frac{7}{2}({nR\Delta\,T})=35 \mathrm{J}$

$1.$ efficiency more than $27 \%$
$2.$ efficiency less than the efficiency a Carnot engine operating between the same two temperatures.
$3.$ efficiency equal to $27 \%$
$4.$ efficiency less than $27 \%$
$I.$ Area $ABCD =$ Work done on the gas
$II.$ Area $ABCD =$ Net heat absorbed
$III.$ Change in the internal energy in cycle $= 0$
Which of these are correct
[ $R$ is the gas constant]
$(1)$ Work done in this thermodynamic cycle $(1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1)$ is $| W |=\frac{1}{2} RT _0$
$(2)$ The ratio of heat transfer during processes $1 \rightarrow 2$ and $2 \rightarrow 3$ is $\left|\frac{ Q _{1 \rightarrow 2}}{ Q _{2 \rightarrow 3}}\right|=\frac{5}{3}$
$(3)$ The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
$(4)$ The ratio of heat transfer during processes $1 \rightarrow 2$ and $3 \rightarrow 4$ is $\left|\frac{Q_{U \rightarrow 2}}{Q_{3 \rightarrow 4}}\right|=\frac{1}{2}$

