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One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the new temperature of the gas
- a)333 K
- b)316 K
- c)343 K
- d)301 K
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
One mole of an ideal monatomic gas is at an initial temperature of 300...
Explanation:
for monoatomic gas CV = 1.5R, CP = 2.5R
At const volume,
Q = 500 J
Q=nCVΔT
500=1×1.5×8.31(T1−300)
T1=340K
At const pressure Q = 500 J
Q=nCPΔT
500=1×2.5×8.31(340−T2)
T2=316K
Most Upvoted Answer
One mole of an ideal monatomic gas is at an initial temperature of 300...
Given:
- Initial temperature of the gas (T1) = 300 K
- Heat added to the gas in an isovolumetric process (Q1) = 500 J
- Heat removed from the gas in an isobaric process (Q2) = -500 J
Concept:
In an isovolumetric process (constant volume), the change in internal energy (ΔU) is equal to the heat added to the gas (Q1). Mathematically, ΔU = Q1.
In an isobaric process (constant pressure), the change in internal energy (ΔU) is given by the equation: ΔU = Q2 + W, where W is the work done by the gas. Since the volume is constant, the work done is zero (W = 0). Therefore, ΔU = Q2.
The change in internal energy (ΔU) of an ideal monatomic gas is given by the equation: ΔU = (3/2) nRΔT, where n is the number of moles of the gas and R is the gas constant.
Solution:
Step 1: Calculate the change in internal energy (ΔU) in the isovolumetric process:
ΔU = Q1 = 500 J
Step 2: Calculate the change in internal energy (ΔU) in the isobaric process:
ΔU = Q2 = -500 J
Step 3: Equate the two expressions for ΔU and solve for the change in temperature (ΔT):
(3/2) nRΔT = 500 J
ΔT = (2/3) * (500 J) / (nR)
Step 4: Substitute the given values:
n = 1 mole (given)
R = 8.314 J/(mol·K) (gas constant)
ΔT = (2/3) * (500 J) / (1 mol * 8.314 J/(mol·K))
Step 5: Simplify the expression:
ΔT = 1000 J / (3 * 8.314 J/(mol·K))
ΔT = 1000 K / (3 * 8.314 K)
ΔT ≈ 40.27 K
Step 6: Calculate the final temperature (T2) by adding the change in temperature (ΔT) to the initial temperature (T1):
T2 = T1 + ΔT
T2 = 300 K + 40.27 K
T2 ≈ 340.27 K
Step 7: Round the final temperature (T2) to the nearest whole number to get the answer:
T2 ≈ 340 K
Therefore, the new temperature of the gas is approximately 340 K, which is closest to option B (316 K).
- Initial temperature of the gas (T1) = 300 K
- Heat added to the gas in an isovolumetric process (Q1) = 500 J
- Heat removed from the gas in an isobaric process (Q2) = -500 J
Concept:
In an isovolumetric process (constant volume), the change in internal energy (ΔU) is equal to the heat added to the gas (Q1). Mathematically, ΔU = Q1.
In an isobaric process (constant pressure), the change in internal energy (ΔU) is given by the equation: ΔU = Q2 + W, where W is the work done by the gas. Since the volume is constant, the work done is zero (W = 0). Therefore, ΔU = Q2.
The change in internal energy (ΔU) of an ideal monatomic gas is given by the equation: ΔU = (3/2) nRΔT, where n is the number of moles of the gas and R is the gas constant.
Solution:
Step 1: Calculate the change in internal energy (ΔU) in the isovolumetric process:
ΔU = Q1 = 500 J
Step 2: Calculate the change in internal energy (ΔU) in the isobaric process:
ΔU = Q2 = -500 J
Step 3: Equate the two expressions for ΔU and solve for the change in temperature (ΔT):
(3/2) nRΔT = 500 J
ΔT = (2/3) * (500 J) / (nR)
Step 4: Substitute the given values:
n = 1 mole (given)
R = 8.314 J/(mol·K) (gas constant)
ΔT = (2/3) * (500 J) / (1 mol * 8.314 J/(mol·K))
Step 5: Simplify the expression:
ΔT = 1000 J / (3 * 8.314 J/(mol·K))
ΔT = 1000 K / (3 * 8.314 K)
ΔT ≈ 40.27 K
Step 6: Calculate the final temperature (T2) by adding the change in temperature (ΔT) to the initial temperature (T1):
T2 = T1 + ΔT
T2 = 300 K + 40.27 K
T2 ≈ 340.27 K
Step 7: Round the final temperature (T2) to the nearest whole number to get the answer:
T2 ≈ 340 K
Therefore, the new temperature of the gas is approximately 340 K, which is closest to option B (316 K).
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