In an isobaric process, heat is supplied to a monoatomic ideal gas. The fraction of heat that goes into mechanical work is

Question

In an isobaric process, heat is supplied to a monoatomic ideal gas. The fraction of heat that goes into mechanical work is (A) 1 (B) (2/3) (C) (3/5) (D) (2/5)

Tardigrade Admin · Accepted Answer

QP=nCPΔ T=(5/2)nRΔ TW=PΔ V=nRΔ T⇒ (W/QP)=(n R Δ T/(5)2 n R Δ T)=(2/5)

Q. In an isobaric process, heat is supplied to a monoatomic ideal gas. The fraction of heat that goes into mechanical work is

$1$

$\frac{2}{3}$

$\frac{3}{5}$

$\frac{2}{5}$

$Q_{P} = n C_{P} Δ T = \frac{5}{2} n R Δ T W = P Δ V = n R Δ T \Rightarrow \frac{W}{Q _{P}} = \frac{n R Δ T}{\frac{5}{2} n R Δ T} = \frac{2}{5}$

Q. In an isobaric process, heat is supplied to a monoatomic ideal gas. The fraction of heat that goes into mechanical work is

1

32​

53​

52​

QP​=nCP​ΔT=25​nRΔTW=PΔV=nRΔT⇒QP​W​=25​nRΔTnRΔT​=52​

$1$

$\frac{2}{3}$

$\frac{3}{5}$

$\frac{2}{5}$

$Q_{P} = n C_{P} Δ T = \frac{5}{2} n R Δ T W = P Δ V = n R Δ T \Rightarrow \frac{W}{Q _{P}} = \frac{n R Δ T}{\frac{5}{2} n R Δ T} = \frac{2}{5}$