An insulated closed piston–cylinder device initially contains 0.3 m3 of carbon dioxide at 200 kPa
and 27°C. A resistance heater inside the cylinder is turned on and supplied heat to the gas. As a
result, the gas expanded by pushing the piston up, until the volume doubled. During this process,
6
the pressure changed according to = 4, in which the constant 6 has units of kPa.m
a) Find the mass of the hydrogen in the tank in kg.
b) Determine the work done by the gas in kJ.

To solve this problem, we can use the ideal gas law and the equation for polytropic process.

The ideal gas law is a fundamental law of physics that describes the behavior of an ideal gas. It relates the pressure, volume, temperature, and number of particles of a gas using the following equation:

PV = nRT

a) First, we need to find the mass of the carbon dioxide in the tank. The ideal gas law is:

PV = mRT

where P is the pressure, V is the volume, m is the mass, R is the universal gas constant, and T is the temperature. Rearranging for the mass, we get:

m = PV / RT

Substituting the given values, we have:

m = (200 kPa)(0.3 m3) / [(0.287 kPam3/kgK)(27°C + 273.15)] = 3.87 kg

So the mass of the carbon dioxide in the tank is 3.87 kg.

b) To determine the work done by the gas during the process, we can use the equation for polytropic process:

P1V1^n = P2V2^n

where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and n is the polytropic index. Substituting the given values, we have:

(200 kPa)(0.3 m3)^n = (4)(0.6 m3)^n

Dividing both sides by (0.3 m3)^n and taking the logarithm of both sides, we get:

log(200) + nlog(0.3) = log(4) + nlog(0.6)

Solving for n, we get:

n = log(4/200) / log(0.6/0.3) ≈ 1.235

Using the polytropic work equation:

W = (P2V2 - P1V1) / (1 - n)

Substituting the given values, we have:

W = [(4 kPa)(0.6 m3) - (200 kPa)(0.3 m3)] / (1 - 1.235) = 233.7 kJ

So the work done by the gas during the process is 233.7 kJ.

Learn more about Gas Law at:

https://brainly.com/question/27009857

#SPJ2

1. What is the productivity rate using cycle time for the following information:
I
Type of Work – Hauling
Average Cycle Time – 35 Minutes
Truck Capacity – 25 Tons
Crew - One Driver
Productivity Factor - 0.85
System Efficiency – 55 Minutes
per
Hour

A 50-mm cube of the graphite fiber reinforced polymer matrix composite material is subjected to 125-kN uniformly distributed compressive force in the direction 2, which is perpendicular to the fiber direction (direction 1). The cube is constrained against expansion in direction 3. Determine:

a. changes in the 50-mm dimensions.
b. stresses required to provide constraints.

The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a particle) is 2 m above the road. Calculate the speed v of the truck.