An ideal monoatomic gas of moles is taken through a cycle consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic diagram. The volume of the gas at , and points are , and , respectively. If the absolute tem

Question

An ideal monoatomic gas of $n$ moles is taken through a cycle $W X Y Z W$ consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic $V - T$ diagram. The volume of the gas at $W$ , $X$ and $Y$ points are $64 {cm}^{3}$ , $125 {cm}^{3}$ and $250 {cm}^{3}$ , respectively. If the absolute temperature of the gas $T_{W}$ at the point $W$ is such that $n R T_{W} = 1 J$ ( $R$ is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path $X Y$ is ______. [2025]

Smart Achievers · Accepted Answer

(1.6)

$Q_{x y} = P_{1} (V_{y} - V_{x}) + n R (T_{y} - T_{x}) \frac{3}{2}$

$Q_{x y} = \frac{5}{2} R (T_{y} - T_{x}) n$

$T_{w} \cdot {(64)}^{\frac{2}{3}} = T_{x} \cdot {(125)}^{\frac{2}{3}}$

$T_{w} \cdot 16 = T_{x} \cdot 25$

$T_{x} = \frac{16}{25} T_{w}$

$\frac{T_{y}}{V_{y}} = \frac{T_{x}}{V_{x}}$

$T_{y} = T_{x} (\frac{250}{125}) = 2 T_{x}$

$Q_{x y} = \frac{5}{2} n R (T_{x}) = \frac{5}{2} n R \cdot \frac{16}{25} T_{w}$ $[∵ n R T_{w} = 1 J given]$

$∴$ $Q_{x y} = \frac{8}{5} = 1.6$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (1.6) Qxy=P1(Vy-Vx)+nR(Ty-Tx)32 Qxy=52R(Ty-Tx)n Tw·(64)23=Tx·(125)23 Tw·16=Tx·25 Tx=1625Tw TyVy=TxVx Ty=Tx(250125)=2Tx Qxy=52nR(Tx)=52nR·1625Tw [∵nRTw=1 J given] ∴ Qxy=85=1.6

Want Us to Email you About Special Offers & Updates?