Q. 14

# The internal energy of a gas is given by U = 1.5 pV. It expands from 100 cm3 to 200 cm3 against a constant pressure of 1.0 × 105 Pa. Calculate the heat absorbed by the gas in the process.

Given

Constant pressure p=1.0×105 Pa

Change in volume ΔV=(200-100)×10-6m3=10-4m3

Internal energy U= 1.5pV

So, change in internal energy ΔU=1.5pΔV

=1.5×1.0×105×10-4

=15J

We know that work done by the gas is given as

ΔW=pΔV=1.0×105×10-4=10J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Therefore,

ΔQ=15+10=25J

heat absorbed by the system=25J.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
JEE Advanced Level Questions of NLM48 mins
Work done in Thermodynamic ProcessFREE Class
Microsporogenesis & Megasporogenesis49 mins
NEET 2021 | Transcription - An important Topic62 mins
Meselson and Stahl experiment49 mins
DNA Fingerprinting42 mins
Interactive Quiz on Molecular basis of Inheritance-02FREE Class
MCQs of Ecology for NEET52 mins
Medelian disorder series: Thalassemia, Sickle cell anaemia & Phenylketonuria60 mins
Different high order thinking questions based on DNA structure39 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses