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One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.
Correct answer is '900'. Can you explain this answer?
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The adiabatic expansion of a monatomic ideal gas is described by the equation:
(P1 * V1^γ) = (P2 * V2^γ)
Where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and γ is the adiabatic index (γ = 5/3 for a monatomic ideal gas).
In this case, the volume increases by a factor of 8, so V2 = 8 * V1.
Let's assume the initial pressure is P1 and the final pressure is P2.
Using the equation above, we can rewrite it as:
(P1 * V1^γ) = (P2 * (8 * V1)^γ)
Simplifying, we get:
P1 * V1^γ = P2 * 8^γ * V1^γ
Since the gas is undergoing an adiabatic expansion, there is no heat transfer, which means Q = 0. According to the first law of thermodynamics, we have:
Q = ΔU + W
Where Q is the heat transfer, ΔU is the change in internal energy, and W is the work done by the gas. Since Q = 0, we have:
ΔU + W = 0
For an adiabatic process, ΔU = C_v * ΔT, where C_v is the molar specific heat capacity at constant volume, and ΔT is the change in temperature.
Since the gas is monatomic, C_v = (3/2)R, where R is the universal gas constant.
ΔU = (3/2)R * ΔT
For an adiabatic process, W = -ΔU, so:
W = -(3/2)R * ΔT
Since ΔU + W = 0, we have:
(3/2)R * ΔT - (3/2)R * ΔT = 0
Therefore, the work done by the gas in an adiabatic process is zero.
From the ideal gas law, P1 * V1 = n * R * T1, where n is the number of moles and T1 is the initial temperature.
Rearranging the equation, we get:
P1 = (n * R * T1) / V1
Using the same equation for the final state, we have:
P2 = (n * R * T2) / V2
Substituting the values, we get:
P2 = (n * R * T2) / (8 * V1)
Since the work done by the gas is zero, the final pressure is equal to the initial pressure:
P1 = P2
Substituting the expressions for P1 and P2, we get:
(n * R * T1) / V1 = (n * R * T2) / (8 * V1)
Simplifying, we get:
T2 = T1 / 8
Plugging in the values, we have:
T2 = 100 K / 8 = 12.5 K
Therefore, the final temperature of the gas is 12.5 K.
(P1 * V1^γ) = (P2 * V2^γ)
Where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and γ is the adiabatic index (γ = 5/3 for a monatomic ideal gas).
In this case, the volume increases by a factor of 8, so V2 = 8 * V1.
Let's assume the initial pressure is P1 and the final pressure is P2.
Using the equation above, we can rewrite it as:
(P1 * V1^γ) = (P2 * (8 * V1)^γ)
Simplifying, we get:
P1 * V1^γ = P2 * 8^γ * V1^γ
Since the gas is undergoing an adiabatic expansion, there is no heat transfer, which means Q = 0. According to the first law of thermodynamics, we have:
Q = ΔU + W
Where Q is the heat transfer, ΔU is the change in internal energy, and W is the work done by the gas. Since Q = 0, we have:
ΔU + W = 0
For an adiabatic process, ΔU = C_v * ΔT, where C_v is the molar specific heat capacity at constant volume, and ΔT is the change in temperature.
Since the gas is monatomic, C_v = (3/2)R, where R is the universal gas constant.
ΔU = (3/2)R * ΔT
For an adiabatic process, W = -ΔU, so:
W = -(3/2)R * ΔT
Since ΔU + W = 0, we have:
(3/2)R * ΔT - (3/2)R * ΔT = 0
Therefore, the work done by the gas in an adiabatic process is zero.
From the ideal gas law, P1 * V1 = n * R * T1, where n is the number of moles and T1 is the initial temperature.
Rearranging the equation, we get:
P1 = (n * R * T1) / V1
Using the same equation for the final state, we have:
P2 = (n * R * T2) / V2
Substituting the values, we get:
P2 = (n * R * T2) / (8 * V1)
Since the work done by the gas is zero, the final pressure is equal to the initial pressure:
P1 = P2
Substituting the expressions for P1 and P2, we get:
(n * R * T1) / V1 = (n * R * T2) / (8 * V1)
Simplifying, we get:
T2 = T1 / 8
Plugging in the values, we have:
T2 = 100 K / 8 = 12.5 K
Therefore, the final temperature of the gas is 12.5 K.
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One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.Correct answer is '900'. Can you explain this answer?
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One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.Correct answer is '900'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.Correct answer is '900'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.Correct answer is '900'. Can you explain this answer?.
One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.Correct answer is '900'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.Correct answer is '900'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.Correct answer is '900'. Can you explain this answer?.
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