6 questions · timed · auto-graded
Work done for process from 1 to 2
$\text{WD}=\int\limits^{\text{v}_{2}}_{\text{v}_{1}}\text{P}.\text{dV}=\int\limits^{\text{v}_{2}}_{\text{v}_{1}}\frac{\text{K}}{\text{V}^\frac{1}{2}}\text{dV}=\text{K}\int\limits^{\text{v}_{2}}_{\text{v}_{1}}\text{V}^{-\big(\frac{1}{2}\big)\text{dV}}$$$
$\text{WD}=\text{K}\Bigg[\frac{\text{V}^\frac{1}{2}}{\frac{1}{2}}\Bigg]^{\text{v}_2}_{\text{v}_1}=2\text{K}\big[\sqrt{\text{V}_{2}}-\sqrt{\text{V}_1}\big]$
WD form $\text{V}_1\text{to}\text{V}_2,\text{i.e},\text{dW}=2\text{P}_1\text{V}_1^\frac{1}{2}\big[\sqrt{\text{V}_2}-\sqrt{\text{V}_1}\big]$
$=2\text{P}_2\text{V}_2^\frac{1}{2}\big[\sqrt{\text{V}_2}-\sqrt{\text{V}_1}\big]$
$5=\frac{\text{T}_{2}}{300-\text{T}_{2}}\Rightarrow\text{T}_{2}=1500-5\text{T}_{2}$
$\text{T}_{2}+5\text{T}_{2}=1500\Rightarrow6\text{T}_{2}=1500$
$\text{T}_{2}=\frac{1500}{6}=250\text{K}=250-273=-23^\circ\text{C}.$
$\therefore\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$
$\Rightarrow\Delta\text{Q}=\Delta\text{W}$ so heat supplied to the system is utilized in expansion system is isothermal.
What is the ratio of temperature $\frac{\text{T}_1}{\text{T}_2},\text{if}\text{V}_2=2\text{V}_1?$from gas equation of ideal gas PV nRT
$\Rightarrow\text{T}=\frac{\text{PV}}{\text{nR}}=\frac{\text{P}\sqrt{\text{V}}\sqrt{\text{V}}}{\text{nR}}=\frac{\text{K}\sqrt{\text{V}}}{\text{nR}}$
$\text{T}_1\frac{\text{K}\sqrt{\text{V}_1}}{\text{nR}}\text{and}\text{T}_2=\frac{\text{K}\sqrt{\text{V}_2}}{\text{nR}}$
$\frac{\text{T}_1}{\text{T}_2}=\frac{\frac{\text{K}\sqrt{\text{V}}_1}{\text{nR}}}{\frac{\text{K}\sqrt{\text{V}}_2}{\text{nR}}}=\frac{\sqrt{\text{V}}_1}{\sqrt{\text{V}}_2}=\sqrt\frac{\text{V}_1}{2\text{V}_1}$ $(\therefore\ \text{V}_2=2\text{V}_1\text{given})$
$\therefore\ \frac{\text{T}_1}{\text{T}_2}=\frac{1}{\sqrt2}\ ....(\text{ii})$
required ratio is $1:\sqrt2$