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When a monatomic ideal gas expands at a constant pressure of 6.4 x 105 Pa, the volume of the gas increases by 3.2 x 10-3 m3. Determine the heat that flows. If heat flows into the gas, then the heat flow is positive. If the heat flows out of the gas, then the heat flow is negative.

Respuesta :

Manetho

Answer:

5118.50 J

Explanation:

pΔv=nRΔT ;

therefore, ΔT=PV/nR

ΔT = (6.4×10^5)(3.2×10^(-3)/1×8.314

ΔT= 2.4633×10^2 = 246.33 K

specific heat at constant pressure is given as:

c_p = 3/2R

c_p = 12.5 J/mol K

Now, substitute in equation (1)

we know that

Q=ΔU+W ;

and

W=pΔV= 6.4×10^5×3.2×10^(-3) = 2048 J

now

ΔU=CvΔT = 12.465×246.33 =3070.50 J ;

therefore  

Q=3070.50+2048= 5118.50 J

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