Answer :
Answer
given,
V = 2 L
the left is an ideal gas at P = 100 k Pa and T = 500 K
mass is constant
[tex]m_1 = m_2[/tex]
[tex]\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}[/tex]
Pressure is same because it's not changing due to process
[tex]\dfrac{V}{500} = \dfrac{2 V}{T_2}[/tex]
[tex]T_2 = 1000\ K[/tex]
[tex]\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}[/tex]
[tex]\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})[/tex]
[tex]m = \dfrac{P_1V_1}{RT_1}[/tex]
[tex]m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}[/tex]
m = 1.39 x 10⁻³ Kg
[tex]\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)[/tex]
[tex]\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K[/tex]