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Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at constant volume, adding 1.36 * 104 J of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. Calculate (a) the final temperature of the gas; (b) the amount of work done by the gas; (c) the amount of heat added to the gas while it was expanding; (d) the change in internal energy of the gas for the whole process.

Respuesta :

okpalawalter8

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]

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