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Answer:
a) [tex]T_b=590.775k[/tex]
b) [tex]W_t=1.08*10^4J[/tex]
d) [tex]Q=3.778*10^4J[/tex]
d) [tex]\triangle V=4.058*10^4J[/tex]
Explanation:
From the question we are told that:
Moles of N2 [tex]n=2.50[/tex]
Atmospheric pressure [tex]P=100atm[/tex]
Temperature [tex]t=20 \textdegree C[/tex]
[tex]t = 20+273[/tex]
[tex]t = 293k[/tex]
Initial heat [tex]Q=1.36 * 10^4 J[/tex]
a)
Generally the equation for change in temperature is mathematically given by
[tex]\triangle T=\frac{Q}{N*C_v}[/tex]
Where
[tex]C_v=Heat\ Capacity \approx 20.76 J/mol/K[/tex]
[tex]T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }[/tex]
[tex]T_b-293k=297.775[/tex]
[tex]T_b=590.775k[/tex]
b)
Generally the equation for ideal gas is mathematically given by
[tex]PV=nRT[/tex]
For v double
[tex]T_c=2*590.775k[/tex]
[tex]T_c=1181.55k[/tex]
Therefore
[tex]PV=Wbc[/tex]
[tex]Wbc=(2.20)(8.314)(1181_590.778)[/tex]
[tex]Wbc=10805.7J[/tex]
Total Work-done [tex]W_t[/tex]
[tex]W_t=Wab+Wbc[/tex]
[tex]W_t=0+1.08*10^4[/tex]
[tex]W_t=1.08*10^4J[/tex]
c)
Generally the equation for amount of heat added is mathematically given by
[tex]Q=nC_p\triangle T[/tex]
[tex]Q=2.20*2907*(1181.55-590.775)\\[/tex]
[tex]Q=3.778*10^4J[/tex]
d)
Generally the equation for change in internal energy of the gas is mathematically given by
[tex]\triangle V=nC_v \triangle T[/tex]
[tex]\triangle V=2.20*20.76*(1181.55-293)k[/tex]
[tex]\triangle V=4.058*10^4J[/tex]