148294
In an isobaric process, the work done by a diatomic gas is $10 \mathrm{~J}$, the heat given to the gas will be:
1 $35 \mathrm{~J}$
2 $30 \mathrm{~J}$
3 $45 \mathrm{~J}$
4 $60 \mathrm{~J}$
Explanation:
A For constant pressure process $\mathrm{W}=10 \mathrm{~J}$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\mathrm{nCp} \Delta \mathrm{T}}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\mathrm{n}\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R} \Delta \mathrm{T}}=\frac{1}{\frac{\mathrm{f}}{2}+1}$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{1}{\frac{5}{2}+1} \quad\left[\begin{array}{l}\mathrm{f}=\text { Degree of freedom } \\ \mathrm{f}=5 \text { for dia atomic gas }\end{array}\right]$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{2}{7}$ $\mathrm{Q}=\frac{7 \mathrm{~W}}{2}$ $\mathrm{Q}=\frac{7 \times 10}{2}$ $\mathrm{Q}=35$ Joule
AIIMS-25.05.2019(E) Shift-2
Thermodynamics
148291
Assertion: Air quickly leaking out of a balloon becomes cooler. Reason: The leaking air undergoes adiabatic expansion.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A When a system under goes a change under the condition that do not exchange of heat takes places between the system and surroundings. Then such a process called adiabatic process. The leaking air of balloon under goes adiabatic expansion. In this expansion due to work done against external pressure, the internal energy of air reduces, Thus, it becomes cooler.
AIIMS-26.05.2019(E) Shift-2
Thermodynamics
148295
A cycle process is shown on p-T diagram. Which of the following curves shows the same process on $\mathrm{p}-\mathrm{V}$ digaram.
1
2
3
4
Explanation:
B $\frac{\mathrm{PV}}{\mathrm{T}}=$ constant $\mathrm{AB}$ is isochoric process, $\mathrm{V}=\mathrm{c}$ $\mathrm{BC}$ is isobaric process, $\mathrm{P}=\mathrm{c}$ When $\mathrm{T}$ is decreasing $\mathrm{v}$ is decreasing. Hence, CA is isothermal process When, $\mathrm{P}$ is decreasing $\mathrm{v}$ is increasing. Hence, diagram will rectangular hyperbola.
BCECE-2016
Thermodynamics
148296
At constant pressure, the ratio of increase in volume of an ideal gas per degree rise in kelvin temperature to its original volume is-
1 $\mathrm{T}^{2}$
2 $\frac{1}{\mathrm{~T}}$
3 $\mathrm{T}^{3}$
4 $\mathrm{T}$
Explanation:
B Ideal gas equation $\mathrm{PV}=\mathrm{RT}$ $\mathrm{V}=\left(\frac{\mathrm{R}}{\mathrm{P}}\right) \mathrm{T}$ $\mathrm{V}$, is directly proportional to $\mathrm{T}$. $\mathrm{V} \propto \mathrm{T}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}-1=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}-1$ [By subtracting $\mathrm{V}_{1}$ from both side] $\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}}=\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1}} \quad\left[\mathrm{~T}_{2}-\mathrm{T}_{1}=1 \mathrm{~K}\right]$ $\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}}=\frac{1}{\mathrm{~T}_{1}}=\frac{1}{\mathrm{~T}}$
148294
In an isobaric process, the work done by a diatomic gas is $10 \mathrm{~J}$, the heat given to the gas will be:
1 $35 \mathrm{~J}$
2 $30 \mathrm{~J}$
3 $45 \mathrm{~J}$
4 $60 \mathrm{~J}$
Explanation:
A For constant pressure process $\mathrm{W}=10 \mathrm{~J}$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\mathrm{nCp} \Delta \mathrm{T}}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\mathrm{n}\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R} \Delta \mathrm{T}}=\frac{1}{\frac{\mathrm{f}}{2}+1}$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{1}{\frac{5}{2}+1} \quad\left[\begin{array}{l}\mathrm{f}=\text { Degree of freedom } \\ \mathrm{f}=5 \text { for dia atomic gas }\end{array}\right]$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{2}{7}$ $\mathrm{Q}=\frac{7 \mathrm{~W}}{2}$ $\mathrm{Q}=\frac{7 \times 10}{2}$ $\mathrm{Q}=35$ Joule
AIIMS-25.05.2019(E) Shift-2
Thermodynamics
148291
Assertion: Air quickly leaking out of a balloon becomes cooler. Reason: The leaking air undergoes adiabatic expansion.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A When a system under goes a change under the condition that do not exchange of heat takes places between the system and surroundings. Then such a process called adiabatic process. The leaking air of balloon under goes adiabatic expansion. In this expansion due to work done against external pressure, the internal energy of air reduces, Thus, it becomes cooler.
AIIMS-26.05.2019(E) Shift-2
Thermodynamics
148295
A cycle process is shown on p-T diagram. Which of the following curves shows the same process on $\mathrm{p}-\mathrm{V}$ digaram.
1
2
3
4
Explanation:
B $\frac{\mathrm{PV}}{\mathrm{T}}=$ constant $\mathrm{AB}$ is isochoric process, $\mathrm{V}=\mathrm{c}$ $\mathrm{BC}$ is isobaric process, $\mathrm{P}=\mathrm{c}$ When $\mathrm{T}$ is decreasing $\mathrm{v}$ is decreasing. Hence, CA is isothermal process When, $\mathrm{P}$ is decreasing $\mathrm{v}$ is increasing. Hence, diagram will rectangular hyperbola.
BCECE-2016
Thermodynamics
148296
At constant pressure, the ratio of increase in volume of an ideal gas per degree rise in kelvin temperature to its original volume is-
1 $\mathrm{T}^{2}$
2 $\frac{1}{\mathrm{~T}}$
3 $\mathrm{T}^{3}$
4 $\mathrm{T}$
Explanation:
B Ideal gas equation $\mathrm{PV}=\mathrm{RT}$ $\mathrm{V}=\left(\frac{\mathrm{R}}{\mathrm{P}}\right) \mathrm{T}$ $\mathrm{V}$, is directly proportional to $\mathrm{T}$. $\mathrm{V} \propto \mathrm{T}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}-1=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}-1$ [By subtracting $\mathrm{V}_{1}$ from both side] $\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}}=\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1}} \quad\left[\mathrm{~T}_{2}-\mathrm{T}_{1}=1 \mathrm{~K}\right]$ $\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}}=\frac{1}{\mathrm{~T}_{1}}=\frac{1}{\mathrm{~T}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
148294
In an isobaric process, the work done by a diatomic gas is $10 \mathrm{~J}$, the heat given to the gas will be:
1 $35 \mathrm{~J}$
2 $30 \mathrm{~J}$
3 $45 \mathrm{~J}$
4 $60 \mathrm{~J}$
Explanation:
A For constant pressure process $\mathrm{W}=10 \mathrm{~J}$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\mathrm{nCp} \Delta \mathrm{T}}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\mathrm{n}\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R} \Delta \mathrm{T}}=\frac{1}{\frac{\mathrm{f}}{2}+1}$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{1}{\frac{5}{2}+1} \quad\left[\begin{array}{l}\mathrm{f}=\text { Degree of freedom } \\ \mathrm{f}=5 \text { for dia atomic gas }\end{array}\right]$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{2}{7}$ $\mathrm{Q}=\frac{7 \mathrm{~W}}{2}$ $\mathrm{Q}=\frac{7 \times 10}{2}$ $\mathrm{Q}=35$ Joule
AIIMS-25.05.2019(E) Shift-2
Thermodynamics
148291
Assertion: Air quickly leaking out of a balloon becomes cooler. Reason: The leaking air undergoes adiabatic expansion.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A When a system under goes a change under the condition that do not exchange of heat takes places between the system and surroundings. Then such a process called adiabatic process. The leaking air of balloon under goes adiabatic expansion. In this expansion due to work done against external pressure, the internal energy of air reduces, Thus, it becomes cooler.
AIIMS-26.05.2019(E) Shift-2
Thermodynamics
148295
A cycle process is shown on p-T diagram. Which of the following curves shows the same process on $\mathrm{p}-\mathrm{V}$ digaram.
1
2
3
4
Explanation:
B $\frac{\mathrm{PV}}{\mathrm{T}}=$ constant $\mathrm{AB}$ is isochoric process, $\mathrm{V}=\mathrm{c}$ $\mathrm{BC}$ is isobaric process, $\mathrm{P}=\mathrm{c}$ When $\mathrm{T}$ is decreasing $\mathrm{v}$ is decreasing. Hence, CA is isothermal process When, $\mathrm{P}$ is decreasing $\mathrm{v}$ is increasing. Hence, diagram will rectangular hyperbola.
BCECE-2016
Thermodynamics
148296
At constant pressure, the ratio of increase in volume of an ideal gas per degree rise in kelvin temperature to its original volume is-
1 $\mathrm{T}^{2}$
2 $\frac{1}{\mathrm{~T}}$
3 $\mathrm{T}^{3}$
4 $\mathrm{T}$
Explanation:
B Ideal gas equation $\mathrm{PV}=\mathrm{RT}$ $\mathrm{V}=\left(\frac{\mathrm{R}}{\mathrm{P}}\right) \mathrm{T}$ $\mathrm{V}$, is directly proportional to $\mathrm{T}$. $\mathrm{V} \propto \mathrm{T}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}-1=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}-1$ [By subtracting $\mathrm{V}_{1}$ from both side] $\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}}=\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1}} \quad\left[\mathrm{~T}_{2}-\mathrm{T}_{1}=1 \mathrm{~K}\right]$ $\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}}=\frac{1}{\mathrm{~T}_{1}}=\frac{1}{\mathrm{~T}}$
148294
In an isobaric process, the work done by a diatomic gas is $10 \mathrm{~J}$, the heat given to the gas will be:
1 $35 \mathrm{~J}$
2 $30 \mathrm{~J}$
3 $45 \mathrm{~J}$
4 $60 \mathrm{~J}$
Explanation:
A For constant pressure process $\mathrm{W}=10 \mathrm{~J}$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\mathrm{nCp} \Delta \mathrm{T}}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\mathrm{n}\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R} \Delta \mathrm{T}}=\frac{1}{\frac{\mathrm{f}}{2}+1}$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{1}{\frac{5}{2}+1} \quad\left[\begin{array}{l}\mathrm{f}=\text { Degree of freedom } \\ \mathrm{f}=5 \text { for dia atomic gas }\end{array}\right]$ $\frac{\mathrm{W}}{\mathrm{Q}}=\frac{2}{7}$ $\mathrm{Q}=\frac{7 \mathrm{~W}}{2}$ $\mathrm{Q}=\frac{7 \times 10}{2}$ $\mathrm{Q}=35$ Joule
AIIMS-25.05.2019(E) Shift-2
Thermodynamics
148291
Assertion: Air quickly leaking out of a balloon becomes cooler. Reason: The leaking air undergoes adiabatic expansion.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A When a system under goes a change under the condition that do not exchange of heat takes places between the system and surroundings. Then such a process called adiabatic process. The leaking air of balloon under goes adiabatic expansion. In this expansion due to work done against external pressure, the internal energy of air reduces, Thus, it becomes cooler.
AIIMS-26.05.2019(E) Shift-2
Thermodynamics
148295
A cycle process is shown on p-T diagram. Which of the following curves shows the same process on $\mathrm{p}-\mathrm{V}$ digaram.
1
2
3
4
Explanation:
B $\frac{\mathrm{PV}}{\mathrm{T}}=$ constant $\mathrm{AB}$ is isochoric process, $\mathrm{V}=\mathrm{c}$ $\mathrm{BC}$ is isobaric process, $\mathrm{P}=\mathrm{c}$ When $\mathrm{T}$ is decreasing $\mathrm{v}$ is decreasing. Hence, CA is isothermal process When, $\mathrm{P}$ is decreasing $\mathrm{v}$ is increasing. Hence, diagram will rectangular hyperbola.
BCECE-2016
Thermodynamics
148296
At constant pressure, the ratio of increase in volume of an ideal gas per degree rise in kelvin temperature to its original volume is-
1 $\mathrm{T}^{2}$
2 $\frac{1}{\mathrm{~T}}$
3 $\mathrm{T}^{3}$
4 $\mathrm{T}$
Explanation:
B Ideal gas equation $\mathrm{PV}=\mathrm{RT}$ $\mathrm{V}=\left(\frac{\mathrm{R}}{\mathrm{P}}\right) \mathrm{T}$ $\mathrm{V}$, is directly proportional to $\mathrm{T}$. $\mathrm{V} \propto \mathrm{T}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}-1=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}-1$ [By subtracting $\mathrm{V}_{1}$ from both side] $\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}}=\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1}} \quad\left[\mathrm{~T}_{2}-\mathrm{T}_{1}=1 \mathrm{~K}\right]$ $\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}}=\frac{1}{\mathrm{~T}_{1}}=\frac{1}{\mathrm{~T}}$
