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The pV diagram shows a process abc involving 0.690 mole of an ideal gas.

A) What was the temperature of this gas at point a? Express your answer in kelvin to three significant figures.

B) What was the temperature of this gas at point b? Express your answer in kelvin to three significant figures.

C) What was the temperature of this gas at point c? Express your answer in kelvin to three significant figures.

D) How much work was done by the gas in this process? Express your answer in joules to two significant figures.

E) How much heat had to be put in during the process to increase the internal energy of the fast by 81000 J? Express your answer in joules to three significant figures.

The PV Diagram Shows A Process Abc Involving 0690 Mole Of An Ideal Gas A What Was The Temperature Of This Gas At Point A Express Your Answer In Kelvin To Three class=

Sagot :

MathPhys

Answer:

A) 349 K

B) 6.10×10³ K

C) 9.76×10³ K

D) 21,000 J

E) 102,000 J

Explanation:

A-C) The ideal gas law states PV = nRT, where:

  • P is pressure in Pa
  • V is volume in m³
  • n is number of moles
  • R is the universal gas constant, 8.314 J/kg/K
  • T is temperature in K

Given the number of moles, and the pressure and volume at each point, we can solve for temperature.

A) At point a, P = 2.0×10⁵ Pa and V = 0.010 m³. The temperature is:

PV = nRT

(2.0×10⁵ Pa) (0.010 m³) = (0.690 mol) (8.314 J/kg/K) T

T = 349 K

B) At point b, P = 5.0×10⁵ Pa and V = 0.070 m³. The temperature is:

PV = nRT

(5.0×10⁵ Pa) (0.070 m³) = (0.690 mol) (8.314 J/kg/K) T

T = 6.10×10³ K

C) At point c, P = 8.0×10⁵ Pa and V = 0.070 m³. The temperature is:

PV = nRT

(8.0×10⁵ Pa) (0.070 m³) = (0.690 mol) (8.314 J/kg/K) T

T = 9.76×10³ K

D) The magnitude of the work is equal to the area under the P-V curve. In this case, the region is a trapezoid.

W = ½ (2.0×10⁵ Pa + 5.0×10⁵ Pa) (0.070 m³ − 0.010 m³)

W = 21,000 J

Since the volume is increasing, the work done to the gas is negative. The work done by the gas is positive.

E) From the first law of thermodynamics, the sum of the work done to the gas (W) and the heat added to it (Q) is equal to the net increase in internal energy (ΔU).

W + Q = ΔU

Given that the work done to the gas is W = -21,000 J, and the increase in internal energy is ΔU = 81,000 J, then the heat added is:

-21,000 J + Q = 81,000 J

Q = 102,000 J

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