371478
In which of the following processes, heat is neither absorbed nor released by a system?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
In adiabatic process, heat exchange does not take place .
NEET - 2019
PHXI12:THERMODYNAMICS
371479
Assertion : An adiabatic process is an isoentropic process. Reason : Change in entropy is zero in case of adiabatic process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In adiabatic process \(\Delta S=\dfrac{\Delta Q}{T}=0\), since there is no heat exchange \((\Delta Q=0)\) (with the surroundings). \(\Rightarrow S=\) constant So, this process is called isentropic So correct option is (1).
PHXI12:THERMODYNAMICS
371480
Assertion : Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion, adiabatically. Reason : Temperature remains constant in isothermal expansion and not in adiabatic expansion.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Adiabatic \(\Delta Q=0\) \(\Rightarrow|\Delta U|=|\Delta W|\) (by first law of thermodynamics) Isothermal \(\Delta T=0\) \(\Rightarrow \Delta Q=\Delta W\) \(\Rightarrow\) complete \(\Delta Q\) is used as work in isothermal. \(\Rightarrow\) work done in isothermal expansion is greater than in adiabatic expansion. However, the reason lacks precision, attributing the difference to temperature alone, while it's influenced by the change in internal energy. So correct option is (2).
PHXI12:THERMODYNAMICS
371481
\(\Delta U+\Delta W=0\) is valid for
1 Isothermal process
2 Adiabatic process
3 Isochoric process
4 Isobaric process
Explanation:
In adiabatic process \(\begin{aligned}& \Delta Q=0 \Rightarrow \Delta U+\Delta W=0 \\& (\because \Delta Q=\Delta U+\Delta W)\end{aligned}\)
PHXI12:THERMODYNAMICS
371482
A tyre pumped to a pressure \(3.375 \mathrm{~atm}\) at \(27^\circ C\) suddenly bursts. What is the final temperature \((\gamma=1.5)\) ?
1 \( - 27^\circ C\)
2 \(27^\circ C\)
3 \( - 73^\circ C\)
4 \(0^\circ C\)
Explanation:
Sudden bursting of a tyre can be treated as adiabatic expansion \(T_{1}^{\gamma} P_{1}^{1-\gamma}=T_{2}^{\gamma} P_{2}^{1-\gamma}\) \(\left(\dfrac{T_{1}}{T_{2}}\right)^{\gamma}=\left(\dfrac{P_{1}}{P_{2}}\right)^{\gamma-1} \Rightarrow\left(\dfrac{300}{T_{2}}\right)^{3 / 2}=\left(\dfrac{3.375}{1}\right)^{3 / 2-1}\) \({T_2} = \frac{{300}}{{{{(3.375)}^{1/3}}}} = 200\;K = - 73^\circ C\)
371478
In which of the following processes, heat is neither absorbed nor released by a system?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
In adiabatic process, heat exchange does not take place .
NEET - 2019
PHXI12:THERMODYNAMICS
371479
Assertion : An adiabatic process is an isoentropic process. Reason : Change in entropy is zero in case of adiabatic process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In adiabatic process \(\Delta S=\dfrac{\Delta Q}{T}=0\), since there is no heat exchange \((\Delta Q=0)\) (with the surroundings). \(\Rightarrow S=\) constant So, this process is called isentropic So correct option is (1).
PHXI12:THERMODYNAMICS
371480
Assertion : Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion, adiabatically. Reason : Temperature remains constant in isothermal expansion and not in adiabatic expansion.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Adiabatic \(\Delta Q=0\) \(\Rightarrow|\Delta U|=|\Delta W|\) (by first law of thermodynamics) Isothermal \(\Delta T=0\) \(\Rightarrow \Delta Q=\Delta W\) \(\Rightarrow\) complete \(\Delta Q\) is used as work in isothermal. \(\Rightarrow\) work done in isothermal expansion is greater than in adiabatic expansion. However, the reason lacks precision, attributing the difference to temperature alone, while it's influenced by the change in internal energy. So correct option is (2).
PHXI12:THERMODYNAMICS
371481
\(\Delta U+\Delta W=0\) is valid for
1 Isothermal process
2 Adiabatic process
3 Isochoric process
4 Isobaric process
Explanation:
In adiabatic process \(\begin{aligned}& \Delta Q=0 \Rightarrow \Delta U+\Delta W=0 \\& (\because \Delta Q=\Delta U+\Delta W)\end{aligned}\)
PHXI12:THERMODYNAMICS
371482
A tyre pumped to a pressure \(3.375 \mathrm{~atm}\) at \(27^\circ C\) suddenly bursts. What is the final temperature \((\gamma=1.5)\) ?
1 \( - 27^\circ C\)
2 \(27^\circ C\)
3 \( - 73^\circ C\)
4 \(0^\circ C\)
Explanation:
Sudden bursting of a tyre can be treated as adiabatic expansion \(T_{1}^{\gamma} P_{1}^{1-\gamma}=T_{2}^{\gamma} P_{2}^{1-\gamma}\) \(\left(\dfrac{T_{1}}{T_{2}}\right)^{\gamma}=\left(\dfrac{P_{1}}{P_{2}}\right)^{\gamma-1} \Rightarrow\left(\dfrac{300}{T_{2}}\right)^{3 / 2}=\left(\dfrac{3.375}{1}\right)^{3 / 2-1}\) \({T_2} = \frac{{300}}{{{{(3.375)}^{1/3}}}} = 200\;K = - 73^\circ C\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI12:THERMODYNAMICS
371478
In which of the following processes, heat is neither absorbed nor released by a system?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
In adiabatic process, heat exchange does not take place .
NEET - 2019
PHXI12:THERMODYNAMICS
371479
Assertion : An adiabatic process is an isoentropic process. Reason : Change in entropy is zero in case of adiabatic process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In adiabatic process \(\Delta S=\dfrac{\Delta Q}{T}=0\), since there is no heat exchange \((\Delta Q=0)\) (with the surroundings). \(\Rightarrow S=\) constant So, this process is called isentropic So correct option is (1).
PHXI12:THERMODYNAMICS
371480
Assertion : Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion, adiabatically. Reason : Temperature remains constant in isothermal expansion and not in adiabatic expansion.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Adiabatic \(\Delta Q=0\) \(\Rightarrow|\Delta U|=|\Delta W|\) (by first law of thermodynamics) Isothermal \(\Delta T=0\) \(\Rightarrow \Delta Q=\Delta W\) \(\Rightarrow\) complete \(\Delta Q\) is used as work in isothermal. \(\Rightarrow\) work done in isothermal expansion is greater than in adiabatic expansion. However, the reason lacks precision, attributing the difference to temperature alone, while it's influenced by the change in internal energy. So correct option is (2).
PHXI12:THERMODYNAMICS
371481
\(\Delta U+\Delta W=0\) is valid for
1 Isothermal process
2 Adiabatic process
3 Isochoric process
4 Isobaric process
Explanation:
In adiabatic process \(\begin{aligned}& \Delta Q=0 \Rightarrow \Delta U+\Delta W=0 \\& (\because \Delta Q=\Delta U+\Delta W)\end{aligned}\)
PHXI12:THERMODYNAMICS
371482
A tyre pumped to a pressure \(3.375 \mathrm{~atm}\) at \(27^\circ C\) suddenly bursts. What is the final temperature \((\gamma=1.5)\) ?
1 \( - 27^\circ C\)
2 \(27^\circ C\)
3 \( - 73^\circ C\)
4 \(0^\circ C\)
Explanation:
Sudden bursting of a tyre can be treated as adiabatic expansion \(T_{1}^{\gamma} P_{1}^{1-\gamma}=T_{2}^{\gamma} P_{2}^{1-\gamma}\) \(\left(\dfrac{T_{1}}{T_{2}}\right)^{\gamma}=\left(\dfrac{P_{1}}{P_{2}}\right)^{\gamma-1} \Rightarrow\left(\dfrac{300}{T_{2}}\right)^{3 / 2}=\left(\dfrac{3.375}{1}\right)^{3 / 2-1}\) \({T_2} = \frac{{300}}{{{{(3.375)}^{1/3}}}} = 200\;K = - 73^\circ C\)
371478
In which of the following processes, heat is neither absorbed nor released by a system?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
In adiabatic process, heat exchange does not take place .
NEET - 2019
PHXI12:THERMODYNAMICS
371479
Assertion : An adiabatic process is an isoentropic process. Reason : Change in entropy is zero in case of adiabatic process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In adiabatic process \(\Delta S=\dfrac{\Delta Q}{T}=0\), since there is no heat exchange \((\Delta Q=0)\) (with the surroundings). \(\Rightarrow S=\) constant So, this process is called isentropic So correct option is (1).
PHXI12:THERMODYNAMICS
371480
Assertion : Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion, adiabatically. Reason : Temperature remains constant in isothermal expansion and not in adiabatic expansion.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Adiabatic \(\Delta Q=0\) \(\Rightarrow|\Delta U|=|\Delta W|\) (by first law of thermodynamics) Isothermal \(\Delta T=0\) \(\Rightarrow \Delta Q=\Delta W\) \(\Rightarrow\) complete \(\Delta Q\) is used as work in isothermal. \(\Rightarrow\) work done in isothermal expansion is greater than in adiabatic expansion. However, the reason lacks precision, attributing the difference to temperature alone, while it's influenced by the change in internal energy. So correct option is (2).
PHXI12:THERMODYNAMICS
371481
\(\Delta U+\Delta W=0\) is valid for
1 Isothermal process
2 Adiabatic process
3 Isochoric process
4 Isobaric process
Explanation:
In adiabatic process \(\begin{aligned}& \Delta Q=0 \Rightarrow \Delta U+\Delta W=0 \\& (\because \Delta Q=\Delta U+\Delta W)\end{aligned}\)
PHXI12:THERMODYNAMICS
371482
A tyre pumped to a pressure \(3.375 \mathrm{~atm}\) at \(27^\circ C\) suddenly bursts. What is the final temperature \((\gamma=1.5)\) ?
1 \( - 27^\circ C\)
2 \(27^\circ C\)
3 \( - 73^\circ C\)
4 \(0^\circ C\)
Explanation:
Sudden bursting of a tyre can be treated as adiabatic expansion \(T_{1}^{\gamma} P_{1}^{1-\gamma}=T_{2}^{\gamma} P_{2}^{1-\gamma}\) \(\left(\dfrac{T_{1}}{T_{2}}\right)^{\gamma}=\left(\dfrac{P_{1}}{P_{2}}\right)^{\gamma-1} \Rightarrow\left(\dfrac{300}{T_{2}}\right)^{3 / 2}=\left(\dfrac{3.375}{1}\right)^{3 / 2-1}\) \({T_2} = \frac{{300}}{{{{(3.375)}^{1/3}}}} = 200\;K = - 73^\circ C\)
371478
In which of the following processes, heat is neither absorbed nor released by a system?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
In adiabatic process, heat exchange does not take place .
NEET - 2019
PHXI12:THERMODYNAMICS
371479
Assertion : An adiabatic process is an isoentropic process. Reason : Change in entropy is zero in case of adiabatic process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In adiabatic process \(\Delta S=\dfrac{\Delta Q}{T}=0\), since there is no heat exchange \((\Delta Q=0)\) (with the surroundings). \(\Rightarrow S=\) constant So, this process is called isentropic So correct option is (1).
PHXI12:THERMODYNAMICS
371480
Assertion : Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion, adiabatically. Reason : Temperature remains constant in isothermal expansion and not in adiabatic expansion.
1 Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Adiabatic \(\Delta Q=0\) \(\Rightarrow|\Delta U|=|\Delta W|\) (by first law of thermodynamics) Isothermal \(\Delta T=0\) \(\Rightarrow \Delta Q=\Delta W\) \(\Rightarrow\) complete \(\Delta Q\) is used as work in isothermal. \(\Rightarrow\) work done in isothermal expansion is greater than in adiabatic expansion. However, the reason lacks precision, attributing the difference to temperature alone, while it's influenced by the change in internal energy. So correct option is (2).
PHXI12:THERMODYNAMICS
371481
\(\Delta U+\Delta W=0\) is valid for
1 Isothermal process
2 Adiabatic process
3 Isochoric process
4 Isobaric process
Explanation:
In adiabatic process \(\begin{aligned}& \Delta Q=0 \Rightarrow \Delta U+\Delta W=0 \\& (\because \Delta Q=\Delta U+\Delta W)\end{aligned}\)
PHXI12:THERMODYNAMICS
371482
A tyre pumped to a pressure \(3.375 \mathrm{~atm}\) at \(27^\circ C\) suddenly bursts. What is the final temperature \((\gamma=1.5)\) ?
1 \( - 27^\circ C\)
2 \(27^\circ C\)
3 \( - 73^\circ C\)
4 \(0^\circ C\)
Explanation:
Sudden bursting of a tyre can be treated as adiabatic expansion \(T_{1}^{\gamma} P_{1}^{1-\gamma}=T_{2}^{\gamma} P_{2}^{1-\gamma}\) \(\left(\dfrac{T_{1}}{T_{2}}\right)^{\gamma}=\left(\dfrac{P_{1}}{P_{2}}\right)^{\gamma-1} \Rightarrow\left(\dfrac{300}{T_{2}}\right)^{3 / 2}=\left(\dfrac{3.375}{1}\right)^{3 / 2-1}\) \({T_2} = \frac{{300}}{{{{(3.375)}^{1/3}}}} = 200\;K = - 73^\circ C\)