Question 456: Thermodynamics & KTG

Question

The above $p$-$v$ diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is
![](https://imagex.cdn.examgoal.net/ikaZL1Ae21HD6Okqd/i6IfZSRFqqq25ltfCXtq4xlBb1LPo/tmFMfIIqU8nDbxBegSWmcj/image.png)

Accepted Answer

Along path DA, volume is constant.
Hence, $\Delta $Q
DA
= nC
v
$\Delta $T = nC
v
(T
A
Ã¢ÂÂ T
D
)
$ 	herefore $ $\Delta $Q
DA
= $n\left( {{3 \over 2}R} ight)\left[ {{{2{p_0}{v_0}} \over {nR}} - {{{p_0}{v_0}} \over {nR}}} ight] = {3 \over 2}{p_0}{v_0}$
Along the path AB, pressure is constant.
Hence $\Delta $Q
AB
= nC
p
$\Delta $T = nC
p
(T
B
Ã¢ÂÂ T
A
)
$ 	herefore $ $\Delta $Q
AB
= $n\left( {{5 \over 2}R} ight)\left[ {{{2{p_0}2{v_0}} \over {nR}} - {{2{p_0}{v_0}} \over {nR}}} ight] = {{10} \over 2}{p_0}{v_0}$
$ 	herefore $  The amount of heat extracted from the source in a single cycle is
$\Delta $Q = $\Delta $Q
DA
+ $\Delta $Q
AB
$ = {3 \over 2}{p_0}{v_0} + {{10} \over 2}{p_0}{v_0}$ = $\left( {{{13} \over 2}} ight){p_0}{v_0}$

Answer

${p_0}{v_0}$

Answer

$\left( {{{13} \over 2}} \right){p_0}{v_0}$

Answer

$\left( {{{11} \over 2}} \right){p_0}{v_0}$

Answer

$4{p_0}{v_0}$

Thermodynamics & KTG

The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is

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