Question 683: Thermodynamics & KTG

Question

When a system is taken from state ' $a$ ' to state ' $c$ ' along a path abc, it is found that $\mathrm{Q}=80 \mathrm{cal}$ and $\mathrm{W}=35 \mathrm{cal}$. Along path adc $\mathrm{Q}=65 \mathrm{cal}$ the work done W along path adc is

Accepted Answer

When a system transitions from state 'a' to state 'c', the change in internal energy, denoted as $\Delta U$, remains constant regardless of the path taken.
For the path
abc
, the heat absorbed $\mathrm{Q}$ is 80 cal and the work done $\mathrm{W}$ is 35 cal. Therefore, the change in internal energy is calculated as:
$   \Delta \mathrm{U} = \mathrm{Q} - \mathrm{W} = 80 \, 	ext{cal} - 35 \, 	ext{cal} = 45 \, 	ext{cal}   $
For the path
adc
, we know $\Delta U = 45 \, 	ext{cal}$ and the heat absorbed is $\mathrm{Q} = 65 \, 	ext{cal}$.
To find the work done along path adc ($\mathrm{W}$), we use the formula:
$ \mathrm{W} = \mathrm{Q} - \Delta \mathrm{U} = 65 \, 	ext{cal} - 45 \, 	ext{cal} = 20 \, 	ext{cal} $

Answer

20 cal.

Answer

35 cal.

Answer

45 cal.

Answer

65 cal.

Thermodynamics & KTG

When a system is taken from state ' a ' to state ' c ' along a path abc, it is found that \mathrm{Q}=80 \mathrm{cal} and \mathrm{W}=35 \mathrm{cal}. Along path adc \mathrm{Q}=65 \mathrm{cal} the work done W along path adc is

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