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A monatomic ideal gas that is initially at a pressure of 1.50 x 10^5 Pa and has a volume of 0.08 m^3 is compressed adiabatically to a volume of 0.0400 m^3.

(a) What is the final pressure?
(b) How much work is done by the gas?
(c) What is the ratio of the final temperature of the gas to its initial temperature? Is the gas heated or cooled by this compression?

User Tymspy
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2 Answers

5 votes

Final answer:

The final pressure is 3.00 x 10^5 Pa. The work done by the gas is -3000 J. The ratio of the final temperature to the initial temperature is approximately 1.25.

Step-by-step explanation:

(a) Final Pressure:

To find the final pressure, we can use Boyle's law, which states that the product of the initial pressure and volume is equal to the product of the final pressure and volume:

P1V1 = P2V2

Given that the initial pressure (P1) is 1.50 x 105 Pa, the initial volume (V1) is 0.08 m3, and the final volume (V2) is 0.0400 m3, we can substitute these values into the equation:

(1.50 x 105 Pa)(0.08 m3) = (P2)(0.0400 m3)

Solving for P2, we find:

P2 = (1.50 x 105 Pa)(0.08 m3)/(0.0400 m3)

P2 = 3.00 x 105 Pa

Therefore, the final pressure is 3.00 x 105 Pa.

(b) Work Done by the Gas:

The work done by the gas during an adiabatic compression can be calculated using the following formula:

Work = -(P2V2 - P1V1)/(y - 1)

where y is the ratio of specific heat capacities, which is 5/3 for a monatomic gas.

Substituting the given values:

Work = -[(3.00 x 105 Pa)(0.0400 m3) - (1.50 x 105 Pa)(0.08 m3)]/(5/3 - 1)

Work = -3000 J

Therefore, the gas does -3000 J of work.

(c) Ratio of Final Temperature to Initial Temperature:

The ratio of the final temperature (T2) to the initial temperature (T1) can be determined using the following formula for an adiabatic compression:

(T2/T1) = (V1/V2)^((y-1)/y)

Substituting the given values into the equation:

(T2/300 K) = (0.08 m3/0.0400 m3)^((5/3-1)/(5/3))

(T2/300 K) = (2)^((2/3)/(5/3))

(T2/300 K) = (2)^(2/5)

T2 ≈ 374 K

Therefore, the ratio of the final temperature to the initial temperature is approximately 374 K / 300 K = 1.25. The gas is heated by this compression.

User Daniel Jennings
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5.4k points
7 votes

Answer:

final pressure is
4.77 * 10^5 \ Pa.


W=-1.05 * 10^4 \ J.


(T_2)/(T_1)=1.59. and gas is heated.

Step-by-step explanation:

We know for mono atomic gas ,
\gamma=1.67.

We know for adiabatic process,


P_1V_1^\gamma=P_2V_2^\gamma.

Putting all values we get :

P_2=
4.77 * 10^5 \ Pa

Therefore final pressure is
4.77 * 10^5 \ Pa.

Now,

Work Done for adiabatic process is ,
W= (1)/(\gamma-1)* (P_1V_1-P_2V_2).

Putting all given values we get:


W=-1.05 * 10^4 \ J.

Now,

We know ,
T_1V_1^(\gamma-1)=T_2V_2^(\gamma-1).

Putting all given values , we get.


(T_2)/(T_1)=1.59..

SInce the ratio of temperature of final and initial temperature of gas is 1.59.

So, Final temperature is more.

Therefore, Gas is heated.

User Sagar Gautam
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4.3k points