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Q. What is the amount of work done when two moles of ideal gas is compressed from a volume of $1 m^{3}$ to $10 \,dm^{3}$ at $300 \,K $ against a pressure of $100 kPa$ ?

MHT CETMHT CET 2015

Solution:

Given, number of moles $=2$ moles

$V_{1}=1 m^{3}=1000 dm ^{3}$

$V_{2}=10 dm ^{3}$

$\therefore $ We have, work done $(W)= -p_{ext} d V$

$=-100 kPa \times(10-1000) dm ^{3}$

$=-100 kPa \times-999 dm ^{3}$

$=99900 J$

$=99.9 kJ$