JEE MAIN - Chemistry (2019 - 12th April Morning Slot - No. 12)
An ideal gas is allowed to expand form 1 L to 10 L against a constant external pressure of I bar. The work
done in kJ is :
+10.0
–0.9
– 2.0
– 9.0
Explanation
This is an irreverseable process as gas is expanding against a constant external process.
Work done in irreverseable process
W = - Pext$$\Delta $$V
= - 1 bar $$ \times $$ 9 L
= - 105 Pa $$ \times $$ 9 $$ \times $$ 10-3 m3
= - 9 $$ \times $$ 102 N-m
= - 900 J
= - 0.9 kJ
Work done in irreverseable process
W = - Pext$$\Delta $$V
= - 1 bar $$ \times $$ 9 L
= - 105 Pa $$ \times $$ 9 $$ \times $$ 10-3 m3
= - 9 $$ \times $$ 102 N-m
= - 900 J
= - 0.9 kJ
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