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A container holds 1.1 g of argon at a pressure of 8.0 atm . you may want to review (pages 382 - 383) . how much will the temperature increase if this amount of heat energy is transferred to the gas at constant pressure? express your answer using two significant figures.

Respuesta :

superman1987
When heat energy at a constant pressure = the change of enthalpy

ΔH = Q

and when the internal energy of an ideal gas formula is:

ΔH = n.Cp.ΔT

and when Cp = (5/2) R   at a constant pressure of a monatomic ideal gas

∴Q = n.(5/2).R.ΔT

∴ ΔT = Q / [n.(5/2).R]

when Q is the heat required = 43.08 J (given)

and n is no.of moles = mass / molar mass

                                  = 1.1 g / 39.948 g/mol

                                  =  0.0275 moles

and R is gas constant = 8.314472J/mol K

by substitution:

ΔT = 43.08 J / (0.0275moles *(5/2) * 8.314472 J/mol K

      = 76.76  K 

The change in temperature of the gas is 60.02°C

Data;

  • Mass = 1.1g
  • Pressure = 8.0 atm

The Number of Moles

The numbers of moles present is needed to calculate the Cv value of the gas.

[tex]n = \frac{mass}{molar mass}\\n = \frac{1.1}{39.9}\\ n = 0.027 moles[/tex]

The Cv value of the gas is

[tex]C_v = \frac{3}{2}R\\ C_v = \frac{3}{2} * 8.314 = 12.471 J/mol[/tex]

The energy of the gas will be

[tex]Q = nC_v\delta T\\Q = 0.027 *12.471 * 100\\Q = 33.67J[/tex]

The energy on the gas is 33.67J

However,

[tex]Q = nC_p \delta T\\[/tex]

But Cp is given as

[tex]C_p = \frac{5}{2} R\\[/tex]

Let's substitute the values and solve

[tex]Q = nC_p \delta T\\33.67 = 0.027 * \frac{5}{2} * 8.314 * \delta T\\33.67 = 0.561\delta T\\\delta T = \frac{33.67}0.561} \\\delta T = 60.02^0C[/tex]

The change in temperature of the gas is 60.02°C

Learn more on Cv, Cp value of a gas here;

https://brainly.com/question/13719615

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