If one mole monoatomic ideal gas was taken through process AB as shown in figure, then select correct option(s)
Given : ln 1.5 = 0.4

Question

If one mole monoatomic ideal gas was taken through process AB as shown in figure, then select correct option(s) Given : ln 1.5 = 0.4

TestHub · Accepted Answer

Process Equation: The line equation for the $V - T$ graph is $V = \frac{T}{60} + 5$ , which gives pressure $P = \frac{RT}{V} = 60 R - \frac{300 R}{V}$ .

Work Done ( $w$ ): $w = - \int_{10}^{15} P d V = - [300 R - 300 R ln (1.5)] = - 180 \times 8.314 = - 1496.52J$ (Option A is correct).

Internal Energy ( $Δ U$ ): $Δ U = n C_{v} Δ T = 1 \times \frac{3}{2} R \times 300 = 450 \times 8.314 = 3741.3 J$ .

Heat ( $q$ ) & Entropy ( $Δ S$ ): $q = Δ U - w = 3741.3 - (- 1496.52) = 5237.82J$ (Option B is correct), and $Δ S$ is positive because both $T$ and $V$ increase (Option D is correct).

Correct Options: A, B, D

Chemistry - Thermodynamics - I Question with Solution | TestHub

Options:(select one or more)

Doubts & Discussion