371551
The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is:
1 \(\dfrac{2}{5}\)
2 \(\dfrac{3}{2}\)
3 \(\dfrac{3}{5}\)
4 \(\dfrac{2}{3}\)
Explanation:
Efficiency of heat engine is given by \(\eta=\dfrac{W}{Q}=\dfrac{P \Delta V}{n C_{P} \Delta T}=\dfrac{n R \Delta T}{n C_{P} \Delta T}=\dfrac{R}{C_{P}}=\dfrac{R}{\dfrac{5 R}{2}}=\dfrac{2}{5}\) For monoatomic gas \(C_{P}=\dfrac{5}{2} R\).
JEE - 2016
PHXI12:THERMODYNAMICS
371552
A gas is compressed at a constant pressure of \(50\;N{\rm{/}}{m^2}\) from a volume of \(10\;{m^3}\) to volume of \(4\;{m^3}\). Energy of \(100\,joules\) is thus added to gas by heating. Its internal energy is:
1 \(450\;J\)
2 \(400\;J\)
3 \(350\;J\)
4 \(410\;J\)
Explanation:
As volume decreases, work is done on the gas and so \(-v e\). \(W=50[4-10]\) \( = - 300{\rm{ }}Joule\) Heat supplied, \(Q = 100{\rm{ }}Joule\) \(Q=\Delta U+W \quad\) (\({{\rm{1}}^{{\rm{st}}}}\) Law of thermodynamics) \(\Delta U=Q-W\) \(=100-(-300)\) \( = 400{\rm{ }}Joule.\)
PHXI12:THERMODYNAMICS
371553
In an isochoric if \({T_1} = 27^\circ C\) and \({T_2} = 127^\circ C\), then \(P_{1} / P_{2}\) will be equal to
1 \(9 / 59\)
2 \(2 / 3\)
3 \(3 / 4\)
4 None of these.
Explanation:
At constant volume \(P \propto T \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{T_{1}}{T_{2}} \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{300}{400}=\dfrac{3}{4}\).
PHXI12:THERMODYNAMICS
371554
What amount of heat must be supplied to \(35\;g\) of oxygen at room temperature to raise its temperature. by \(80^\circ C\) at constant volume (molecular mass of oxygen is 32 and \(R = 8.3\;J\;mo{l^{ - 1}}{k^{ - 1}}\))
1 \(1.52\;kJ\)
2 \(3.23\;kJ\)
3 \(1.81\;kJ\)
4 \(1.62\;kJ\)
Explanation:
Here mass of oxygen \((\mathrm{m})=35 \mathrm{~g}\), molar mass of \(\mathrm{O}_{2}(\mathrm{M})=32 \mathrm{~g} \mathrm{~mol}^{-1}\) rise in temperature, \(\Delta T = 80^\circ C\) \(\therefore\) number of moles \(n = \frac{m}{M} = \frac{{35}}{{32}} = 1.09\;mol\) As oxygen is a diatomic gas, then molar specific heat at constant volume is \(C_{v}=\dfrac{5}{2} R\) and amount of heat supplied to gas \(Q = n{C_v}\Delta T\) \( = 1.09 \times \frac{5}{2}R \times 80 = 1.09 \times \frac{5}{2} \times 8.3 \times 80\) \( = 1.81\;kJ\)
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PHXI12:THERMODYNAMICS
371551
The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is:
1 \(\dfrac{2}{5}\)
2 \(\dfrac{3}{2}\)
3 \(\dfrac{3}{5}\)
4 \(\dfrac{2}{3}\)
Explanation:
Efficiency of heat engine is given by \(\eta=\dfrac{W}{Q}=\dfrac{P \Delta V}{n C_{P} \Delta T}=\dfrac{n R \Delta T}{n C_{P} \Delta T}=\dfrac{R}{C_{P}}=\dfrac{R}{\dfrac{5 R}{2}}=\dfrac{2}{5}\) For monoatomic gas \(C_{P}=\dfrac{5}{2} R\).
JEE - 2016
PHXI12:THERMODYNAMICS
371552
A gas is compressed at a constant pressure of \(50\;N{\rm{/}}{m^2}\) from a volume of \(10\;{m^3}\) to volume of \(4\;{m^3}\). Energy of \(100\,joules\) is thus added to gas by heating. Its internal energy is:
1 \(450\;J\)
2 \(400\;J\)
3 \(350\;J\)
4 \(410\;J\)
Explanation:
As volume decreases, work is done on the gas and so \(-v e\). \(W=50[4-10]\) \( = - 300{\rm{ }}Joule\) Heat supplied, \(Q = 100{\rm{ }}Joule\) \(Q=\Delta U+W \quad\) (\({{\rm{1}}^{{\rm{st}}}}\) Law of thermodynamics) \(\Delta U=Q-W\) \(=100-(-300)\) \( = 400{\rm{ }}Joule.\)
PHXI12:THERMODYNAMICS
371553
In an isochoric if \({T_1} = 27^\circ C\) and \({T_2} = 127^\circ C\), then \(P_{1} / P_{2}\) will be equal to
1 \(9 / 59\)
2 \(2 / 3\)
3 \(3 / 4\)
4 None of these.
Explanation:
At constant volume \(P \propto T \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{T_{1}}{T_{2}} \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{300}{400}=\dfrac{3}{4}\).
PHXI12:THERMODYNAMICS
371554
What amount of heat must be supplied to \(35\;g\) of oxygen at room temperature to raise its temperature. by \(80^\circ C\) at constant volume (molecular mass of oxygen is 32 and \(R = 8.3\;J\;mo{l^{ - 1}}{k^{ - 1}}\))
1 \(1.52\;kJ\)
2 \(3.23\;kJ\)
3 \(1.81\;kJ\)
4 \(1.62\;kJ\)
Explanation:
Here mass of oxygen \((\mathrm{m})=35 \mathrm{~g}\), molar mass of \(\mathrm{O}_{2}(\mathrm{M})=32 \mathrm{~g} \mathrm{~mol}^{-1}\) rise in temperature, \(\Delta T = 80^\circ C\) \(\therefore\) number of moles \(n = \frac{m}{M} = \frac{{35}}{{32}} = 1.09\;mol\) As oxygen is a diatomic gas, then molar specific heat at constant volume is \(C_{v}=\dfrac{5}{2} R\) and amount of heat supplied to gas \(Q = n{C_v}\Delta T\) \( = 1.09 \times \frac{5}{2}R \times 80 = 1.09 \times \frac{5}{2} \times 8.3 \times 80\) \( = 1.81\;kJ\)
371551
The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is:
1 \(\dfrac{2}{5}\)
2 \(\dfrac{3}{2}\)
3 \(\dfrac{3}{5}\)
4 \(\dfrac{2}{3}\)
Explanation:
Efficiency of heat engine is given by \(\eta=\dfrac{W}{Q}=\dfrac{P \Delta V}{n C_{P} \Delta T}=\dfrac{n R \Delta T}{n C_{P} \Delta T}=\dfrac{R}{C_{P}}=\dfrac{R}{\dfrac{5 R}{2}}=\dfrac{2}{5}\) For monoatomic gas \(C_{P}=\dfrac{5}{2} R\).
JEE - 2016
PHXI12:THERMODYNAMICS
371552
A gas is compressed at a constant pressure of \(50\;N{\rm{/}}{m^2}\) from a volume of \(10\;{m^3}\) to volume of \(4\;{m^3}\). Energy of \(100\,joules\) is thus added to gas by heating. Its internal energy is:
1 \(450\;J\)
2 \(400\;J\)
3 \(350\;J\)
4 \(410\;J\)
Explanation:
As volume decreases, work is done on the gas and so \(-v e\). \(W=50[4-10]\) \( = - 300{\rm{ }}Joule\) Heat supplied, \(Q = 100{\rm{ }}Joule\) \(Q=\Delta U+W \quad\) (\({{\rm{1}}^{{\rm{st}}}}\) Law of thermodynamics) \(\Delta U=Q-W\) \(=100-(-300)\) \( = 400{\rm{ }}Joule.\)
PHXI12:THERMODYNAMICS
371553
In an isochoric if \({T_1} = 27^\circ C\) and \({T_2} = 127^\circ C\), then \(P_{1} / P_{2}\) will be equal to
1 \(9 / 59\)
2 \(2 / 3\)
3 \(3 / 4\)
4 None of these.
Explanation:
At constant volume \(P \propto T \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{T_{1}}{T_{2}} \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{300}{400}=\dfrac{3}{4}\).
PHXI12:THERMODYNAMICS
371554
What amount of heat must be supplied to \(35\;g\) of oxygen at room temperature to raise its temperature. by \(80^\circ C\) at constant volume (molecular mass of oxygen is 32 and \(R = 8.3\;J\;mo{l^{ - 1}}{k^{ - 1}}\))
1 \(1.52\;kJ\)
2 \(3.23\;kJ\)
3 \(1.81\;kJ\)
4 \(1.62\;kJ\)
Explanation:
Here mass of oxygen \((\mathrm{m})=35 \mathrm{~g}\), molar mass of \(\mathrm{O}_{2}(\mathrm{M})=32 \mathrm{~g} \mathrm{~mol}^{-1}\) rise in temperature, \(\Delta T = 80^\circ C\) \(\therefore\) number of moles \(n = \frac{m}{M} = \frac{{35}}{{32}} = 1.09\;mol\) As oxygen is a diatomic gas, then molar specific heat at constant volume is \(C_{v}=\dfrac{5}{2} R\) and amount of heat supplied to gas \(Q = n{C_v}\Delta T\) \( = 1.09 \times \frac{5}{2}R \times 80 = 1.09 \times \frac{5}{2} \times 8.3 \times 80\) \( = 1.81\;kJ\)
371551
The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is:
1 \(\dfrac{2}{5}\)
2 \(\dfrac{3}{2}\)
3 \(\dfrac{3}{5}\)
4 \(\dfrac{2}{3}\)
Explanation:
Efficiency of heat engine is given by \(\eta=\dfrac{W}{Q}=\dfrac{P \Delta V}{n C_{P} \Delta T}=\dfrac{n R \Delta T}{n C_{P} \Delta T}=\dfrac{R}{C_{P}}=\dfrac{R}{\dfrac{5 R}{2}}=\dfrac{2}{5}\) For monoatomic gas \(C_{P}=\dfrac{5}{2} R\).
JEE - 2016
PHXI12:THERMODYNAMICS
371552
A gas is compressed at a constant pressure of \(50\;N{\rm{/}}{m^2}\) from a volume of \(10\;{m^3}\) to volume of \(4\;{m^3}\). Energy of \(100\,joules\) is thus added to gas by heating. Its internal energy is:
1 \(450\;J\)
2 \(400\;J\)
3 \(350\;J\)
4 \(410\;J\)
Explanation:
As volume decreases, work is done on the gas and so \(-v e\). \(W=50[4-10]\) \( = - 300{\rm{ }}Joule\) Heat supplied, \(Q = 100{\rm{ }}Joule\) \(Q=\Delta U+W \quad\) (\({{\rm{1}}^{{\rm{st}}}}\) Law of thermodynamics) \(\Delta U=Q-W\) \(=100-(-300)\) \( = 400{\rm{ }}Joule.\)
PHXI12:THERMODYNAMICS
371553
In an isochoric if \({T_1} = 27^\circ C\) and \({T_2} = 127^\circ C\), then \(P_{1} / P_{2}\) will be equal to
1 \(9 / 59\)
2 \(2 / 3\)
3 \(3 / 4\)
4 None of these.
Explanation:
At constant volume \(P \propto T \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{T_{1}}{T_{2}} \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{300}{400}=\dfrac{3}{4}\).
PHXI12:THERMODYNAMICS
371554
What amount of heat must be supplied to \(35\;g\) of oxygen at room temperature to raise its temperature. by \(80^\circ C\) at constant volume (molecular mass of oxygen is 32 and \(R = 8.3\;J\;mo{l^{ - 1}}{k^{ - 1}}\))
1 \(1.52\;kJ\)
2 \(3.23\;kJ\)
3 \(1.81\;kJ\)
4 \(1.62\;kJ\)
Explanation:
Here mass of oxygen \((\mathrm{m})=35 \mathrm{~g}\), molar mass of \(\mathrm{O}_{2}(\mathrm{M})=32 \mathrm{~g} \mathrm{~mol}^{-1}\) rise in temperature, \(\Delta T = 80^\circ C\) \(\therefore\) number of moles \(n = \frac{m}{M} = \frac{{35}}{{32}} = 1.09\;mol\) As oxygen is a diatomic gas, then molar specific heat at constant volume is \(C_{v}=\dfrac{5}{2} R\) and amount of heat supplied to gas \(Q = n{C_v}\Delta T\) \( = 1.09 \times \frac{5}{2}R \times 80 = 1.09 \times \frac{5}{2} \times 8.3 \times 80\) \( = 1.81\;kJ\)
