Answered

IDNLearn.com is designed to help you find the answers you need quickly and easily. Get prompt and accurate answers to your questions from our community of knowledgeable experts.

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.

Sagot :

Okpalawalter8idn

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]

Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.