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Physics : asked on nakin45
 14.01.2023

An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to volume 1450 cm3. If the expansion is isothermal, what are (a) the final pressure and (b) the work done by the gas

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24.06.2023, solved by verified expert
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Ranjana Bhatt
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a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

An ideal monatomic gas initially has a temperature, №17886499, 14.01.2023 12:03

An ideal monatomic gas initially has a temperature, №17886499, 14.01.2023 12:03

An ideal monatomic gas initially has a temperature, №17886499, 14.01.2023 12:03

Where:

An ideal monatomic gas initially has a temperature, №17886499, 14.01.2023 12:03: is the initial pressure = 5.79 atm

An ideal monatomic gas initially has a temperature, №17886499, 14.01.2023 12:03: is the final pressure =?

An ideal monatomic gas initially has a temperature, №17886499, 14.01.2023 12:03: is the initial volume = 420 cm³

An ideal monatomic gas initially has a temperature, №17886499, 14.01.2023 12:03: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

An ideal monatomic gas initially has a temperature, №17886499, 14.01.2023 12:03

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

An ideal monatomic gas initially has a temperature, №17886499, 14.01.2023 12:03

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

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Physics
An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to volume 1450 cm3. If the expansion is isothermal, what are (a) the final pressure and (b) the work done by the gas
Step-by-step answer
P Answered by PhD

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

T = \frac{PV}{nR}

T_{i} = T_{f}

\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}

Where:

P_{i}: is the initial pressure = 5.79 atm

P_{f}: is the final pressure =?

V_{i}: is the initial volume = 420 cm³

V_{f}: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

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