A monatomic gas supplied the heat Q very slowly keeping the pressure constant. The work done by the gas will be:

Question

A monatomic gas supplied the heat Q very slowly keeping the pressure constant. The work done by the gas will be: (A)  (2/3)Q  (B)  (3/5)Q  (C)  (2/5)Q  (D)  (1/5)Q

Tardigrade Admin · Accepted Answer

Apply first law of thermodynamic In isobaric process

Q = n C_{P} Δ T

Also,

d U = n C_{V} Δ T

∴ \frac{Q}{d U} = \frac{n C _{P} Δ T}{n C _{V} Δ T} = \frac{C _{P}}{C _{V}} = γ

For monoatomic gas,

γ = \frac{5}{3}

∴ \frac{Q}{d U} = \frac{5}{3}

\Rightarrow d U = \frac{3}{5} Q

Now, form first law of thermodynamics

Q = d U + W

\Rightarrow W = Q - d U

= Q - \frac{3}{5} Q

∴ W = \frac{2}{5} Q

Q. A monatomic gas supplied the heat $Q$ very slowly keeping the pressure constant. The work done by the gas will be:

$\frac{2}{3} Q$

$\frac{3}{5} Q$

$\frac{2}{5} Q$

$\frac{1}{5} Q$

Q. A monatomic gas supplied the heat Q very slowly keeping the pressure constant. The work done by the gas will be:

32​Q

53​Q

52​Q

51​Q

Apply first law of thermodynamic In isobaric process Q=nCP​ΔT Also, dU=nCV​ΔT ∴dUQ​=nCV​ΔTnCP​ΔT​=CV​CP​​=γ For monoatomic gas, γ=35​ ∴dUQ​=35​ ⇒dU=53​Q Now, form first law of thermodynamics Q=dU+W ⇒W=Q−dU =Q−53​Q ∴W=52​Q

Q. A monatomic gas supplied the heat $Q$ very slowly keeping the pressure constant. The work done by the gas will be:

$\frac{2}{3} Q$

$\frac{3}{5} Q$

$\frac{2}{5} Q$

$\frac{1}{5} Q$