Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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.

Posted by

jitender.kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Mains 2025 session 1 registration deadline on November 22; apply on jeemain.nta.nic.in

anam-khan

BTech Admission 2025 Live: JEE Mains, SRMJEEE, MET registrations underway; MHT CET syllabus updates

anam-khan

Centre issues coaching guidelines, prohibits centres from using false claims like '100% selection'

Latest Question