How much energy would neon gas absorb to increase the temperature

[truckerron1]

i have 3 questions that i need help with

Consider 180 g of neon gas. How much energy would it have to absorb to experience a temperature increase of 12 K?

Construct a PV diagram for each of the following. (a) A gas begins with a volume of 14 L under pressure of 3 atm. The gas first undergoes an isobaric compression to a volume of 6 L. Next, the gas experiences an isothermal expansion back to a volume of 14 L. (b). A gas undergoes a cyclic process consisting of two isochoric processes and two isobaric processes.

A particular engine having efficiency 27% operates at temperature 420C. What is its highest possible exhaust temperature?

thanks ron

 

[Astronuc]

One needs to show one's work before requesting help.

However, some hints:

1. The energy is related to the mass, the specific heat and the change in temperature. An example should be found in one's textbook.

2a. P-V diagrams are fairly straight forward. The problem starts with a known state point, P, V. Isobaric means 'constant' pressure, and on is given the new volume of 6L. Isothermal is constant temperature, so one needs to find the relationship of P and V for an isothermal expansion.
Isobaric process - http://www.pha.jhu.edu/~broholm/l35/node4.html
Isothermal process - http://www.pha.jhu.edu/~broholm/l35/node3.html

2b. An isochoric process, also called an isometric process, is a thermodynamic process in which the volume stays constant; ΔV = 0.
See - http://en.wikipedia.org/wiki/Isochoric_process
and - http://www.pha.jhu.edu/~broholm/l35/node2.html

3. Find the definition of Carnot efficiency as a function to the hot and cold temperatures, T_hot and T_cold, respectively.

 

[kyle8921]

As a scientist, my response would be as follows:

1. To determine the amount of energy needed to increase the temperature of 180 g of neon gas by 12 K, we would need to use the specific heat capacity of neon. This value is 1.030 J/gK, which means that 180 g of neon would require 1.030 x 180 x 12 = 22104 J of energy to increase its temperature by 12 K.

2. For the first PV diagram, we would see a rectangular shape with the x-axis representing volume and the y-axis representing pressure. The initial point would be at (14 L, 3 atm) and the gas would undergo an isobaric compression to (6 L, 3 atm). From there, it would experience an isothermal expansion back to (14 L, 3 atm). The work done in this process would be represented by the area under the curve on the PV diagram.

For the second PV diagram, we would see a shape with two horizontal lines (representing the isochoric processes) and two diagonal lines (representing the isobaric processes). The work done in this process would also be represented by the area under the curve on the PV diagram.

3. To determine the highest possible exhaust temperature of the engine with an efficiency of 27% and operating at a temperature of 420C, we would need to use the Carnot efficiency equation: Efficiency = (T1 - T2)/T1, where T1 is the hot reservoir temperature and T2 is the cold reservoir temperature. Rearranging the equation, we get T2 = T1 - (T1 x Efficiency). Plugging in the values, we get T2 = 420 - (420 x 0.27) = 306C. Therefore, the highest possible exhaust temperature would be 306C.