Question 513: 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

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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