At a constant pressure of 20 Pa a gas is compressed from 10 `m^(3)` to 5 `m^(3)`. Later 100 J of heat is added to the system. The change in internal e

Question

At a constant pressure of 20 Pa a gas is compressed from 10 `m^(3)` to 5 `m^(3)`. Later 100 J of heat is added to the system. The change in internal e

At a constant pressure of 20 Pa a gas is compressed from 10 `m^(3)` to 5 `m^(3)`. Later 100 J of heat is added to the system. The change in internal energy is
A. 200 J
B. 400 J, 260 J
C. 300 J
D. 100 J

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
`DeltaW=P(V_(2)-V_(1))=20(5-10)=-100J`
`DeltaQ=100J`
`DeltaQ=DeltaU+DeltaW`
100=`DeltaU-100implies DeltaU=200J`

4 views Answered 2023-02-05 15:29