148373
In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at $T_{1}=300 \mathrm{~K}$ and $T_{2}=200 \mathrm{~K}$. The value of $V_{A}=2$ unit, $V_{B}=8$ unit, $V_{C}=16$ unit. Find the value of $V_{D}$.
1 4 unit
2 $ \lt 4$ unit
3 $>5$ unit
4 5 unit
Explanation:
A Given, $\mathrm{V}_{\mathrm{A}}=2$ unit, $\mathrm{V}_{\mathrm{B}}=8$ unit, $\mathrm{V}_{\mathrm{C}}=16$ unit $\mathrm{T}_{\mathrm{B}}=\mathrm{T}_{\mathrm{A}}=\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{D}}=\mathrm{T}_{2}=200 \mathrm{~K}$ For adiabatic process, $\mathrm{TV}^{(\gamma-1)}=$ Constant Process (A - D) $\mathrm{T}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^{(\gamma-1)}=\mathrm{T}_{\mathrm{D}} \mathrm{V}_{\mathrm{D}}^{(\gamma-1)}$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{\mathrm{V}_{\mathrm{A}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{D}}}$ Process $(\mathrm{B}-\mathrm{C})$, $\mathrm{T}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^{\gamma-1}=\mathrm{T}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}^{\gamma-1}$ $\mathrm{T}_{1}(8)^{\gamma-1}=\mathrm{T}_{2}(16)^{\gamma-1}$ $\left(\frac{16}{8}\right)^{\gamma-1}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{300}{200}$ $(2)^{\gamma-1}=1.5$ From equation (i), $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=\frac{300}{200}=1.5$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=(2)^{\gamma-1} \Rightarrow \frac{\mathrm{V}_{\mathrm{D}}}{2}=2$ $\mathrm{~V}_{\mathrm{D}}=4 \text { unit }$
BITSAT-2018
Thermodynamics
148374
In the $V$ - $T$ diagram shown in adjoining figure, what is the relation between $P_{1}$ and $P_{2}$ ?
1 $\mathrm{P}_{2}=\mathrm{P}_{1}$
2 $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
3 $\mathrm{P}_{2}>\mathrm{P}_{1}$
4 insufficient data
Explanation:
B In an isobaric process $\mathrm{P}=$ constant $\mathrm{V} \propto \mathrm{T}$ $\mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right) \mathrm{T}$ $\therefore \mathrm{V}-\mathrm{T}$ graph is a straight line with slope $\left(\frac{\mathrm{nR}}{\mathrm{P}}\right)$. Since, $(\text { slope })_{2}>(\text { slope })_{1}$ So, we can say that $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
VITEEE-2013
Thermodynamics
148375
Two moles of a monoatomic gas undergoes an isobaric process. If the gas temperature is increased by $20{ }^{\circ} \mathrm{C}$ then the heat absorbed by the gas is (Take $R=8.3 \mathrm{~J} / \mathrm{K}$. mol)
148376
In which process the $\mathrm{PV}$ indicator diagram is a straight line parallel to volume axis.
1 Isothermal
2 Isobaric
3 Irreversible
4 Adiabatic
Explanation:
B $\mathrm{A}$ isobaric process is a thermodynamic process in which the pressure remains constant. This is usually obtained by allowing the volume to expand or contract in such a way to neutralize any pressure changes that would be caused by heat transfer. Given that the $\mathrm{P}-\mathrm{V}$ indicates diagram shows a straight line parallel to volume axis The slope of the graph $\frac{\Delta \mathrm{P}}{\Delta \mathrm{V}}=0$ $\Delta \mathrm{P}=0$ Hence we can say that pressure remains constant.
MP PMT-2009
Thermodynamics
148377
Two moles of a gas is expanded to double its volume by two different processes. One is isobaric and the other is isothermal. If $w_{1}$ and $w_{\mathbf{2}}$ are the works done respectively, then
C We know that, For Isobaric process, Work done $\left(\mathrm{w}_{1}\right)=\mathrm{PdV}$ $=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ $=\mathrm{P}(2 \mathrm{~V}-\mathrm{V})$ $\mathrm{w}_{1} =\mathrm{PV}=\mathrm{nRT}$ For Isothermal process, $\text { Work done }\left(\mathrm{w}_{2}\right) =\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $=\mathrm{nRT} \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{w}_{2}=\mathrm{nRT} \ln 2$ ....(ii) From equation (i) and (ii) Now, $\mathrm{w}_{2}=\mathrm{w}_{1} \ln 2$
148373
In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at $T_{1}=300 \mathrm{~K}$ and $T_{2}=200 \mathrm{~K}$. The value of $V_{A}=2$ unit, $V_{B}=8$ unit, $V_{C}=16$ unit. Find the value of $V_{D}$.
1 4 unit
2 $ \lt 4$ unit
3 $>5$ unit
4 5 unit
Explanation:
A Given, $\mathrm{V}_{\mathrm{A}}=2$ unit, $\mathrm{V}_{\mathrm{B}}=8$ unit, $\mathrm{V}_{\mathrm{C}}=16$ unit $\mathrm{T}_{\mathrm{B}}=\mathrm{T}_{\mathrm{A}}=\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{D}}=\mathrm{T}_{2}=200 \mathrm{~K}$ For adiabatic process, $\mathrm{TV}^{(\gamma-1)}=$ Constant Process (A - D) $\mathrm{T}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^{(\gamma-1)}=\mathrm{T}_{\mathrm{D}} \mathrm{V}_{\mathrm{D}}^{(\gamma-1)}$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{\mathrm{V}_{\mathrm{A}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{D}}}$ Process $(\mathrm{B}-\mathrm{C})$, $\mathrm{T}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^{\gamma-1}=\mathrm{T}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}^{\gamma-1}$ $\mathrm{T}_{1}(8)^{\gamma-1}=\mathrm{T}_{2}(16)^{\gamma-1}$ $\left(\frac{16}{8}\right)^{\gamma-1}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{300}{200}$ $(2)^{\gamma-1}=1.5$ From equation (i), $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=\frac{300}{200}=1.5$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=(2)^{\gamma-1} \Rightarrow \frac{\mathrm{V}_{\mathrm{D}}}{2}=2$ $\mathrm{~V}_{\mathrm{D}}=4 \text { unit }$
BITSAT-2018
Thermodynamics
148374
In the $V$ - $T$ diagram shown in adjoining figure, what is the relation between $P_{1}$ and $P_{2}$ ?
1 $\mathrm{P}_{2}=\mathrm{P}_{1}$
2 $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
3 $\mathrm{P}_{2}>\mathrm{P}_{1}$
4 insufficient data
Explanation:
B In an isobaric process $\mathrm{P}=$ constant $\mathrm{V} \propto \mathrm{T}$ $\mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right) \mathrm{T}$ $\therefore \mathrm{V}-\mathrm{T}$ graph is a straight line with slope $\left(\frac{\mathrm{nR}}{\mathrm{P}}\right)$. Since, $(\text { slope })_{2}>(\text { slope })_{1}$ So, we can say that $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
VITEEE-2013
Thermodynamics
148375
Two moles of a monoatomic gas undergoes an isobaric process. If the gas temperature is increased by $20{ }^{\circ} \mathrm{C}$ then the heat absorbed by the gas is (Take $R=8.3 \mathrm{~J} / \mathrm{K}$. mol)
148376
In which process the $\mathrm{PV}$ indicator diagram is a straight line parallel to volume axis.
1 Isothermal
2 Isobaric
3 Irreversible
4 Adiabatic
Explanation:
B $\mathrm{A}$ isobaric process is a thermodynamic process in which the pressure remains constant. This is usually obtained by allowing the volume to expand or contract in such a way to neutralize any pressure changes that would be caused by heat transfer. Given that the $\mathrm{P}-\mathrm{V}$ indicates diagram shows a straight line parallel to volume axis The slope of the graph $\frac{\Delta \mathrm{P}}{\Delta \mathrm{V}}=0$ $\Delta \mathrm{P}=0$ Hence we can say that pressure remains constant.
MP PMT-2009
Thermodynamics
148377
Two moles of a gas is expanded to double its volume by two different processes. One is isobaric and the other is isothermal. If $w_{1}$ and $w_{\mathbf{2}}$ are the works done respectively, then
C We know that, For Isobaric process, Work done $\left(\mathrm{w}_{1}\right)=\mathrm{PdV}$ $=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ $=\mathrm{P}(2 \mathrm{~V}-\mathrm{V})$ $\mathrm{w}_{1} =\mathrm{PV}=\mathrm{nRT}$ For Isothermal process, $\text { Work done }\left(\mathrm{w}_{2}\right) =\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $=\mathrm{nRT} \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{w}_{2}=\mathrm{nRT} \ln 2$ ....(ii) From equation (i) and (ii) Now, $\mathrm{w}_{2}=\mathrm{w}_{1} \ln 2$
148373
In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at $T_{1}=300 \mathrm{~K}$ and $T_{2}=200 \mathrm{~K}$. The value of $V_{A}=2$ unit, $V_{B}=8$ unit, $V_{C}=16$ unit. Find the value of $V_{D}$.
1 4 unit
2 $ \lt 4$ unit
3 $>5$ unit
4 5 unit
Explanation:
A Given, $\mathrm{V}_{\mathrm{A}}=2$ unit, $\mathrm{V}_{\mathrm{B}}=8$ unit, $\mathrm{V}_{\mathrm{C}}=16$ unit $\mathrm{T}_{\mathrm{B}}=\mathrm{T}_{\mathrm{A}}=\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{D}}=\mathrm{T}_{2}=200 \mathrm{~K}$ For adiabatic process, $\mathrm{TV}^{(\gamma-1)}=$ Constant Process (A - D) $\mathrm{T}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^{(\gamma-1)}=\mathrm{T}_{\mathrm{D}} \mathrm{V}_{\mathrm{D}}^{(\gamma-1)}$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{\mathrm{V}_{\mathrm{A}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{D}}}$ Process $(\mathrm{B}-\mathrm{C})$, $\mathrm{T}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^{\gamma-1}=\mathrm{T}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}^{\gamma-1}$ $\mathrm{T}_{1}(8)^{\gamma-1}=\mathrm{T}_{2}(16)^{\gamma-1}$ $\left(\frac{16}{8}\right)^{\gamma-1}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{300}{200}$ $(2)^{\gamma-1}=1.5$ From equation (i), $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=\frac{300}{200}=1.5$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=(2)^{\gamma-1} \Rightarrow \frac{\mathrm{V}_{\mathrm{D}}}{2}=2$ $\mathrm{~V}_{\mathrm{D}}=4 \text { unit }$
BITSAT-2018
Thermodynamics
148374
In the $V$ - $T$ diagram shown in adjoining figure, what is the relation between $P_{1}$ and $P_{2}$ ?
1 $\mathrm{P}_{2}=\mathrm{P}_{1}$
2 $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
3 $\mathrm{P}_{2}>\mathrm{P}_{1}$
4 insufficient data
Explanation:
B In an isobaric process $\mathrm{P}=$ constant $\mathrm{V} \propto \mathrm{T}$ $\mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right) \mathrm{T}$ $\therefore \mathrm{V}-\mathrm{T}$ graph is a straight line with slope $\left(\frac{\mathrm{nR}}{\mathrm{P}}\right)$. Since, $(\text { slope })_{2}>(\text { slope })_{1}$ So, we can say that $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
VITEEE-2013
Thermodynamics
148375
Two moles of a monoatomic gas undergoes an isobaric process. If the gas temperature is increased by $20{ }^{\circ} \mathrm{C}$ then the heat absorbed by the gas is (Take $R=8.3 \mathrm{~J} / \mathrm{K}$. mol)
148376
In which process the $\mathrm{PV}$ indicator diagram is a straight line parallel to volume axis.
1 Isothermal
2 Isobaric
3 Irreversible
4 Adiabatic
Explanation:
B $\mathrm{A}$ isobaric process is a thermodynamic process in which the pressure remains constant. This is usually obtained by allowing the volume to expand or contract in such a way to neutralize any pressure changes that would be caused by heat transfer. Given that the $\mathrm{P}-\mathrm{V}$ indicates diagram shows a straight line parallel to volume axis The slope of the graph $\frac{\Delta \mathrm{P}}{\Delta \mathrm{V}}=0$ $\Delta \mathrm{P}=0$ Hence we can say that pressure remains constant.
MP PMT-2009
Thermodynamics
148377
Two moles of a gas is expanded to double its volume by two different processes. One is isobaric and the other is isothermal. If $w_{1}$ and $w_{\mathbf{2}}$ are the works done respectively, then
C We know that, For Isobaric process, Work done $\left(\mathrm{w}_{1}\right)=\mathrm{PdV}$ $=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ $=\mathrm{P}(2 \mathrm{~V}-\mathrm{V})$ $\mathrm{w}_{1} =\mathrm{PV}=\mathrm{nRT}$ For Isothermal process, $\text { Work done }\left(\mathrm{w}_{2}\right) =\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $=\mathrm{nRT} \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{w}_{2}=\mathrm{nRT} \ln 2$ ....(ii) From equation (i) and (ii) Now, $\mathrm{w}_{2}=\mathrm{w}_{1} \ln 2$
148373
In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at $T_{1}=300 \mathrm{~K}$ and $T_{2}=200 \mathrm{~K}$. The value of $V_{A}=2$ unit, $V_{B}=8$ unit, $V_{C}=16$ unit. Find the value of $V_{D}$.
1 4 unit
2 $ \lt 4$ unit
3 $>5$ unit
4 5 unit
Explanation:
A Given, $\mathrm{V}_{\mathrm{A}}=2$ unit, $\mathrm{V}_{\mathrm{B}}=8$ unit, $\mathrm{V}_{\mathrm{C}}=16$ unit $\mathrm{T}_{\mathrm{B}}=\mathrm{T}_{\mathrm{A}}=\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{D}}=\mathrm{T}_{2}=200 \mathrm{~K}$ For adiabatic process, $\mathrm{TV}^{(\gamma-1)}=$ Constant Process (A - D) $\mathrm{T}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^{(\gamma-1)}=\mathrm{T}_{\mathrm{D}} \mathrm{V}_{\mathrm{D}}^{(\gamma-1)}$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{\mathrm{V}_{\mathrm{A}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{D}}}$ Process $(\mathrm{B}-\mathrm{C})$, $\mathrm{T}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^{\gamma-1}=\mathrm{T}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}^{\gamma-1}$ $\mathrm{T}_{1}(8)^{\gamma-1}=\mathrm{T}_{2}(16)^{\gamma-1}$ $\left(\frac{16}{8}\right)^{\gamma-1}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{300}{200}$ $(2)^{\gamma-1}=1.5$ From equation (i), $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=\frac{300}{200}=1.5$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=(2)^{\gamma-1} \Rightarrow \frac{\mathrm{V}_{\mathrm{D}}}{2}=2$ $\mathrm{~V}_{\mathrm{D}}=4 \text { unit }$
BITSAT-2018
Thermodynamics
148374
In the $V$ - $T$ diagram shown in adjoining figure, what is the relation between $P_{1}$ and $P_{2}$ ?
1 $\mathrm{P}_{2}=\mathrm{P}_{1}$
2 $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
3 $\mathrm{P}_{2}>\mathrm{P}_{1}$
4 insufficient data
Explanation:
B In an isobaric process $\mathrm{P}=$ constant $\mathrm{V} \propto \mathrm{T}$ $\mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right) \mathrm{T}$ $\therefore \mathrm{V}-\mathrm{T}$ graph is a straight line with slope $\left(\frac{\mathrm{nR}}{\mathrm{P}}\right)$. Since, $(\text { slope })_{2}>(\text { slope })_{1}$ So, we can say that $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
VITEEE-2013
Thermodynamics
148375
Two moles of a monoatomic gas undergoes an isobaric process. If the gas temperature is increased by $20{ }^{\circ} \mathrm{C}$ then the heat absorbed by the gas is (Take $R=8.3 \mathrm{~J} / \mathrm{K}$. mol)
148376
In which process the $\mathrm{PV}$ indicator diagram is a straight line parallel to volume axis.
1 Isothermal
2 Isobaric
3 Irreversible
4 Adiabatic
Explanation:
B $\mathrm{A}$ isobaric process is a thermodynamic process in which the pressure remains constant. This is usually obtained by allowing the volume to expand or contract in such a way to neutralize any pressure changes that would be caused by heat transfer. Given that the $\mathrm{P}-\mathrm{V}$ indicates diagram shows a straight line parallel to volume axis The slope of the graph $\frac{\Delta \mathrm{P}}{\Delta \mathrm{V}}=0$ $\Delta \mathrm{P}=0$ Hence we can say that pressure remains constant.
MP PMT-2009
Thermodynamics
148377
Two moles of a gas is expanded to double its volume by two different processes. One is isobaric and the other is isothermal. If $w_{1}$ and $w_{\mathbf{2}}$ are the works done respectively, then
C We know that, For Isobaric process, Work done $\left(\mathrm{w}_{1}\right)=\mathrm{PdV}$ $=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ $=\mathrm{P}(2 \mathrm{~V}-\mathrm{V})$ $\mathrm{w}_{1} =\mathrm{PV}=\mathrm{nRT}$ For Isothermal process, $\text { Work done }\left(\mathrm{w}_{2}\right) =\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $=\mathrm{nRT} \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{w}_{2}=\mathrm{nRT} \ln 2$ ....(ii) From equation (i) and (ii) Now, $\mathrm{w}_{2}=\mathrm{w}_{1} \ln 2$
148373
In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at $T_{1}=300 \mathrm{~K}$ and $T_{2}=200 \mathrm{~K}$. The value of $V_{A}=2$ unit, $V_{B}=8$ unit, $V_{C}=16$ unit. Find the value of $V_{D}$.
1 4 unit
2 $ \lt 4$ unit
3 $>5$ unit
4 5 unit
Explanation:
A Given, $\mathrm{V}_{\mathrm{A}}=2$ unit, $\mathrm{V}_{\mathrm{B}}=8$ unit, $\mathrm{V}_{\mathrm{C}}=16$ unit $\mathrm{T}_{\mathrm{B}}=\mathrm{T}_{\mathrm{A}}=\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{D}}=\mathrm{T}_{2}=200 \mathrm{~K}$ For adiabatic process, $\mathrm{TV}^{(\gamma-1)}=$ Constant Process (A - D) $\mathrm{T}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^{(\gamma-1)}=\mathrm{T}_{\mathrm{D}} \mathrm{V}_{\mathrm{D}}^{(\gamma-1)}$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{\mathrm{V}_{\mathrm{A}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{D}}}$ Process $(\mathrm{B}-\mathrm{C})$, $\mathrm{T}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^{\gamma-1}=\mathrm{T}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}^{\gamma-1}$ $\mathrm{T}_{1}(8)^{\gamma-1}=\mathrm{T}_{2}(16)^{\gamma-1}$ $\left(\frac{16}{8}\right)^{\gamma-1}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{300}{200}$ $(2)^{\gamma-1}=1.5$ From equation (i), $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=\frac{300}{200}=1.5$ $\left(\frac{\mathrm{V}_{\mathrm{D}}}{2}\right)^{\gamma-1}=(2)^{\gamma-1} \Rightarrow \frac{\mathrm{V}_{\mathrm{D}}}{2}=2$ $\mathrm{~V}_{\mathrm{D}}=4 \text { unit }$
BITSAT-2018
Thermodynamics
148374
In the $V$ - $T$ diagram shown in adjoining figure, what is the relation between $P_{1}$ and $P_{2}$ ?
1 $\mathrm{P}_{2}=\mathrm{P}_{1}$
2 $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
3 $\mathrm{P}_{2}>\mathrm{P}_{1}$
4 insufficient data
Explanation:
B In an isobaric process $\mathrm{P}=$ constant $\mathrm{V} \propto \mathrm{T}$ $\mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right) \mathrm{T}$ $\therefore \mathrm{V}-\mathrm{T}$ graph is a straight line with slope $\left(\frac{\mathrm{nR}}{\mathrm{P}}\right)$. Since, $(\text { slope })_{2}>(\text { slope })_{1}$ So, we can say that $\mathrm{P}_{2} \lt \mathrm{P}_{1}$
VITEEE-2013
Thermodynamics
148375
Two moles of a monoatomic gas undergoes an isobaric process. If the gas temperature is increased by $20{ }^{\circ} \mathrm{C}$ then the heat absorbed by the gas is (Take $R=8.3 \mathrm{~J} / \mathrm{K}$. mol)
148376
In which process the $\mathrm{PV}$ indicator diagram is a straight line parallel to volume axis.
1 Isothermal
2 Isobaric
3 Irreversible
4 Adiabatic
Explanation:
B $\mathrm{A}$ isobaric process is a thermodynamic process in which the pressure remains constant. This is usually obtained by allowing the volume to expand or contract in such a way to neutralize any pressure changes that would be caused by heat transfer. Given that the $\mathrm{P}-\mathrm{V}$ indicates diagram shows a straight line parallel to volume axis The slope of the graph $\frac{\Delta \mathrm{P}}{\Delta \mathrm{V}}=0$ $\Delta \mathrm{P}=0$ Hence we can say that pressure remains constant.
MP PMT-2009
Thermodynamics
148377
Two moles of a gas is expanded to double its volume by two different processes. One is isobaric and the other is isothermal. If $w_{1}$ and $w_{\mathbf{2}}$ are the works done respectively, then
C We know that, For Isobaric process, Work done $\left(\mathrm{w}_{1}\right)=\mathrm{PdV}$ $=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ $=\mathrm{P}(2 \mathrm{~V}-\mathrm{V})$ $\mathrm{w}_{1} =\mathrm{PV}=\mathrm{nRT}$ For Isothermal process, $\text { Work done }\left(\mathrm{w}_{2}\right) =\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$ $=\mathrm{nRT} \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{w}_{2}=\mathrm{nRT} \ln 2$ ....(ii) From equation (i) and (ii) Now, $\mathrm{w}_{2}=\mathrm{w}_{1} \ln 2$
