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5000J of heat are added to two moles an ideal monoatomic gas, initially at a temperature of 500K, while the gas performs 7500J of work. What is the final temperature of the gas ? (in Kelvin)
Correct answer is '400'. Can you explain this answer?
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Verified Answer
5000J of heat are added to two moles an ideal monoatomic gas, initiall...
ΔU = ΔQ – ΔW = 5000 - 7500
ΔU = –2500 J
⇒ ΔT = –100 K
⇒ T = 500 - 100 = 400 K
The correct answer is: 400
ΔU = –2500 J
⇒ ΔT = –100 K
⇒ T = 500 - 100 = 400 K
The correct answer is: 400
Most Upvoted Answer
5000J of heat are added to two moles an ideal monoatomic gas, initiall...
Given:
- Heat added to the gas (Q) = 5000 J
- Work done by the gas (W) = 7500 J
- Initial temperature of the gas (T₁) = 500 K
- Number of moles of the gas (n) = 2
Key Formulas:
- The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
- For an ideal monoatomic gas, the change in internal energy is given by:
ΔU = (3/2) nR ΔT
where R is the gas constant and ΔT is the change in temperature.
- The work done by the gas is given by:
W = -PΔV
where P is the pressure and ΔV is the change in volume.
- For an ideal gas, the change in volume is related to the change in temperature by:
ΔV = nR ΔT/P
Solution:
We can start by using the first law of thermodynamics to find the change in internal energy:
ΔU = Q - W
ΔU = 5000 J - 7500 J
ΔU = -2500 J
Since the gas is monoatomic, we can use the formula for the change in internal energy for an ideal monoatomic gas:
ΔU = (3/2) nR ΔT
-2500 J = (3/2) (2 mol) (8.314 J/(mol·K)) ΔT
-2500 J = 24.942 J/K ΔT
Simplifying the equation:
ΔT = -100.18 K
The negative sign indicates that the temperature has decreased. Therefore, the final temperature of the gas is:
T₂ = T₁ + ΔT
T₂ = 500 K - 100.18 K
T₂ ≈ 400 K
Therefore, the final temperature of the gas is approximately 400 K.
- Heat added to the gas (Q) = 5000 J
- Work done by the gas (W) = 7500 J
- Initial temperature of the gas (T₁) = 500 K
- Number of moles of the gas (n) = 2
Key Formulas:
- The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
- For an ideal monoatomic gas, the change in internal energy is given by:
ΔU = (3/2) nR ΔT
where R is the gas constant and ΔT is the change in temperature.
- The work done by the gas is given by:
W = -PΔV
where P is the pressure and ΔV is the change in volume.
- For an ideal gas, the change in volume is related to the change in temperature by:
ΔV = nR ΔT/P
Solution:
We can start by using the first law of thermodynamics to find the change in internal energy:
ΔU = Q - W
ΔU = 5000 J - 7500 J
ΔU = -2500 J
Since the gas is monoatomic, we can use the formula for the change in internal energy for an ideal monoatomic gas:
ΔU = (3/2) nR ΔT
-2500 J = (3/2) (2 mol) (8.314 J/(mol·K)) ΔT
-2500 J = 24.942 J/K ΔT
Simplifying the equation:
ΔT = -100.18 K
The negative sign indicates that the temperature has decreased. Therefore, the final temperature of the gas is:
T₂ = T₁ + ΔT
T₂ = 500 K - 100.18 K
T₂ ≈ 400 K
Therefore, the final temperature of the gas is approximately 400 K.
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5000J of heat are added to two moles an ideal monoatomic gas, initially at a temperature of 500K, while the gas performs 7500J of work. What is the final temperature of the gas ? (in Kelvin)Correct answer is '400'. Can you explain this answer?
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