For an ideal gas graph is shown for three processes. Process... | PHYSICS - HEAT, INTERNAL ENERGY AND WORK | SolveeIt | Solveeit

Today, November 17

Question

For an ideal gas graph is shown for three processes. Processes 1, 2, and 3 are respectively –

A

Isochoric, isobaric, adiabatic

B

Isochoric, adiabatic, isobaric

C

Isobaric, adiabatic, isochoric

D

Adiabatic, isobaric, isochoric

Answer

Isochoric, isobaric, adiabatic

Explanation

Solution

Isochoric process dV = 0

W = 0

Isobaric : W = PDV = nRDT

adiabatic : W = $\frac{nR(T_{i} - T_{f})}{\gamma - 1}$

| W | = $\frac{nR\Delta T}{\gamma - 1}$

0 < g – 1 < 1