JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 17)
The ratio of work done by an ideal monoatomic gas to the heat supplied to it
in an isobaric process is :
$${3 \over 5}$$
$${2 \over 3}$$
$${3 \over 2}$$
$${2 \over 5}$$
설명
In an isobaric process,
Heat supplied, Q = n Cp $$\Delta $$ T
Work done, w = nR$$\Delta $$T
$$ \therefore $$ Ratio = $${w \over Q}$$ = $${{nR\Delta T} \over {n{C_p}\Delta T}}$$
= $${R \over {{5 \over 2}R}}$$
= $${2 \over 5}$$
[Cp = $${5 \over 2}$$ R for monoatomic gas]
Heat supplied, Q = n Cp $$\Delta $$ T
Work done, w = nR$$\Delta $$T
$$ \therefore $$ Ratio = $${w \over Q}$$ = $${{nR\Delta T} \over {n{C_p}\Delta T}}$$
= $${R \over {{5 \over 2}R}}$$
= $${2 \over 5}$$
[Cp = $${5 \over 2}$$ R for monoatomic gas]