Answer:
Explanation:
Given
Pressure [tex]P_1=1.01\times 10^5 Pa[/tex]
Volume of air [tex]V=60 m^3[/tex]
Initial Temperature [tex]T_1=289 K[/tex]
[tex]T_2=302 K[/tex]
Initial moles is given by
[tex]PV=nRT[/tex]
[tex]n_1=\frac{P_1V_1}{RT_1}[/tex]
[tex]n_1=\frac{1.01\times 10^5\times 60}{R\cdot 289}[/tex]
when some gas escape out
no of moles is equal to
[tex]n_2=\frac{P_2\times V_2}{R\cdot T_2}[/tex]
[tex]n_2=\frac{1.01\times 10^5\times 60}{R\cdot 302}[/tex]
remaining no of moles [tex]=n_1-n_2[/tex]
[tex]=\frac{1.01\times 10^5\times 60}{R\cdot 289}-\frac{1.01\times 10^5\times 60}{R\cdot 302}[/tex]
[tex]=\frac{1.01\times 10^5\times 60}{R}(\frac{1}{289}-\frac{1}{302})[/tex]
Mass of air escape out
[tex]=(n_1-n_2)\times M[/tex]
[tex]=(\frac{1.01\times 10^5\times 60}{R}(\frac{1}{289}-\frac{1}{302}))\times 28[/tex]
[tex]=25.272\times 10^3\ g[/tex]
[tex]m=25.272\ kg[/tex]
