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An ideal gas undergoes a reversible isobaric process. The initial volume is 40 liters, the initial temperature is $300 \mathrm{~K}$ , and the final temperature is $400 \mathrm{~K}$ . Calculate the heat $\mathrm{(Q)}$ added or removed from the gas during the process. Given the heat capacity at constant pressure $\mathrm{\left(C_p\right)\: is \: 25 \mathrm{~J} / \mathrm{K}.}$

Option: 1
$2300 \mathrm{~J}$

Option: 2
$1300 \mathrm{~J}$

Option: 3
$2500 \mathrm{~J}$

Option: 4
$2400 \mathrm{~J}$

Answers (1)

best_answer

Given data:

Initial volume $\mathrm{\left(V_i\right)=40}$ liters

Initial temperature $\mathrm{\left(T_i\right)=300 \mathrm{~K}}$

Final temperature $\mathrm{\left(T_f\right)=400 \mathrm{~K}}$

Heat capacity at constant pressure $\mathrm{\left(C_p\right)=25 \quad \mathrm{~J} / \mathrm{K}}$

The change in heat $\mathrm{(Q)}$ added or removed from the gas during an isobaric process can be calculated using the equation:

$\mathrm{ Q=\eta C_p \Delta T }$
Substitute the values and calculate $\mathrm{ Q: }$

$\mathrm{ Q=(1 \mathrm{~mol}) \cdot(25 \mathrm{~J} / \mathrm{K}) \cdot(400 \mathrm{~K}-300 \mathrm{~K})=2500 \mathrm{~J} }$

Since the temperature increased, heat $\mathrm{ (Q) }$ is added to the gas.

Therefore, the heat $\mathrm{ (Q) }$ added to the gas during the process is $\mathrm{ 2500 \mathrm{~J}.}$

So, correct option is 3.

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jitender.kumar

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