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Question

A quantity of monatomic ideal gas expands adiabatically from a volume of 2.0 liters to 6.0 liters. If the initial pressure is P0, what is the final pressure? 

Ans: 0.16 P0 

A cylinder containing an ideal gas has a volume of 2.0 m3 and a pressure of 1.0*10^5 Pa at a temperature of 300 K. The cylinder is placed against a metal block that is maintained at 900 K and the gas expands as the pressure remains constant until the temperature of the gas reaches 900 K. The change in internal energy of the gas is +6.0*10^5 How much heat did the gas absorb? 

ANS: 10*10^5 

A turbine takes in 1 000-K steam and exhausts the steam at a temperature of 500 K. What is the maximum theoretical efficiency of this system? 

Ans: 50% 

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1*10^6 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was: 

And: Greatest on Day 1The efficiency of a Carnot engine operating between 100°C and 0°C is most nearly: 

Ans: 27% 

A Carnot engine runs between a hot reservoir at Th and a cold reservoir at Tc. If one of the temperatures is either increased or decreased by 3.5 K, which of the following changes would increase the efficiency by the greatest amount? 

Ans: Decreasing Tc

Answer 1

(1)

The Volume of the gas is v1 =  2.0 ltrs.

Pressure at volume P1 = P0.

Now the volume is increased to v2 =  6.0 ltrs.

Ideal gas expanding adiabatically .

Where γ is 1.67 for monatomic gas

P2 = 0.16 P0

Therefore Pressure at volume v2 = 6 ltrs is 0.16 P0. 

Answer 2

(2)

The volume of the gas V = 2.0 m³

Pressure of the Gas P = 1.0 * 10^5 Pa.

Temperature T = 300 K

We know that Ideal gas Equation

PV = nRT

Where R is gas constant = 8.314 J/(K*mol)

PV = nRT

1*10^5 * 2.0 = n * 8.314 * 300

n = 80.1  moles

Now the cylinder is placed in the metal block.

Final Temperature of gas in cylinder = 900 K.

Pressure of the Gas P = 1.0 * 10^5 Pa.

n = 80.1 moles

We know that Ideal gas Equation

PV = nRT

Where R is gas constant = 8.314 J/(K*mol)

PV = nRT

(10^5) * V = 80.1 * 8.314 * 900

V = 5.99 m³

Volume of the gas at Temperature 900 K is 6 m³

Therefore,

Change in the Internal Energy at Constant Pressure

ΔH=ΔU+PΔV

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ΔH=ΔU+PΔV

Where ΔH is Enthalpy

ΔU is Change in the Internal Energy = 6 * 10^5

P is Pressure

ΔV is change in volume.

ΔH = 6 * 10^5 + (10^5 *( 6 - 2))

ΔH = 6 * 10^5 + (10^5 *4)

ΔH = 10 * 10^5 J

Therefore the heat absorbed by the gas is 10 * 10^5 J.

Answer 3

(3)

The Turbine takes 1000 K Temperature

Input Temperature = 1000 K

The stream exhausted at Temperature of 500K

Output Temperature = 500 K

Efficiency = (Output / Input) * 100

Efficiency = (500 / 1000) * 100

Efficiency = 0.5 * 100

Therefore Efficiency is 50%.

Answer 4

(4)

The Temperature is operating between 100°C to 0°C

then Temperature of the Hot reservoir is Th = 100°C = 273 + 100 = 373 K.

Temperature of the cold reservoir is Tc = 0°C = 273 + 0 = 273 K.

Efficiency = 1 - (Tc/Th)

Efficiency = 1 - (273/373)

Efficiency = 1 - 0.7319

Efficiency = 0.2680 * 100

Efficiency = 26.8 %

Efficiency ≈ 27%

Therefore the efficiency of the Carnot Engine is 27%.

Answer 5

(5)

The Temperature of the Hot reservoir is Th

The Temperature of the Cold reservoir is Tc

Efficiency of the Carnot Engine is

We can clearly Observe that decreasing Th or increasing Tc would decrease the value Efficiency.

But increasing Th or decreasing Tc would Increase the value Efficiency

Therefore increasing Th or decreasing Tc would Increase the value Efficiency.