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10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change is
- a)2.422 JK-1
- b)5.98 JK-1
- c)- 2.422 JK-1
- d)- 5.981 JK-1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pre...
For isobaric process, we have ∆S =nCpln(T2/T1)
T2 = 273+127 = 400K and T1 = 273+227 = 300K
Applying pV = nRT at initial condition,
1×10 = n×0.0821×300
n = 0.40
Applying ∆S =nCpln(T2/T1)
∆S =0.40×5/2R×ln(400/300) = 2.38 JK-1
T2 = 273+127 = 400K and T1 = 273+227 = 300K
Applying pV = nRT at initial condition,
1×10 = n×0.0821×300
n = 0.40
Applying ∆S =nCpln(T2/T1)
∆S =0.40×5/2R×ln(400/300) = 2.38 JK-1
Most Upvoted Answer
10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pre...
°C is compressed to a volume of 5 dm3 while the temperature is kept constant. What will be the new pressure of the gas?
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since the gas is ideal and monoatomic, we know that the number of moles (n) of the gas will remain constant during the compression.
Given:
Initial volume (V1) = 10 dm3
Final volume (V2) = 5 dm3
Initial temperature (T1) = 27°C = 27 + 273.15 K = 300.15 K
Since the temperature is kept constant, T1 = T2 = 300.15 K
Using the ideal gas law equation, we can set up the following equation:
P1V1 = P2V2
Substituting the given values:
P1 * 10 dm3 = P2 * 5 dm3
Simplifying:
P1 = (P2 * 5 dm3) / 10 dm3
P1 = P2 * 0.5
Since the temperature is constant, we can use the equation P1V1/T1 = P2V2/T2:
P1 * 10 dm3 / 300.15 K = P2 * 5 dm3 / 300.15 K
Simplifying:
P1 / 30.015 K = P2 / 60.03 K
Substituting P1 = P2 * 0.5:
P2 * 0.5 / 30.015 K = P2 / 60.03 K
Cross-multiplying:
0.5 * P2 * 60.03 K = P2 * 30.015 K
Simplifying:
30.015 P2 K = 60.03 P2 K
Subtracting 30.015 P2 K from both sides:
29.015 P2 K = 0
Since the pressure (P2) cannot be zero, we have an inconsistency in our equation.
Therefore, there is no valid solution for the new pressure of the gas when it is compressed to a volume of 5 dm3 while the temperature is kept constant.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since the gas is ideal and monoatomic, we know that the number of moles (n) of the gas will remain constant during the compression.
Given:
Initial volume (V1) = 10 dm3
Final volume (V2) = 5 dm3
Initial temperature (T1) = 27°C = 27 + 273.15 K = 300.15 K
Since the temperature is kept constant, T1 = T2 = 300.15 K
Using the ideal gas law equation, we can set up the following equation:
P1V1 = P2V2
Substituting the given values:
P1 * 10 dm3 = P2 * 5 dm3
Simplifying:
P1 = (P2 * 5 dm3) / 10 dm3
P1 = P2 * 0.5
Since the temperature is constant, we can use the equation P1V1/T1 = P2V2/T2:
P1 * 10 dm3 / 300.15 K = P2 * 5 dm3 / 300.15 K
Simplifying:
P1 / 30.015 K = P2 / 60.03 K
Substituting P1 = P2 * 0.5:
P2 * 0.5 / 30.015 K = P2 / 60.03 K
Cross-multiplying:
0.5 * P2 * 60.03 K = P2 * 30.015 K
Simplifying:
30.015 P2 K = 60.03 P2 K
Subtracting 30.015 P2 K from both sides:
29.015 P2 K = 0
Since the pressure (P2) cannot be zero, we have an inconsistency in our equation.
Therefore, there is no valid solution for the new pressure of the gas when it is compressed to a volume of 5 dm3 while the temperature is kept constant.
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Community Answer
10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pre...
∆s =ncpln(t2 /t1)
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10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer?
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10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer?.
10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer?.
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10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer?, a detailed solution for 10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of 10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change isa)2.422 JK-1b)5.98JK-1c)- 2.422JK-1d)- 5.981JK-1Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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