371228
A thermodynamical system is changed from state \(\left(p_{1} V_{1}\right)\) to \(\left(p_{2} V_{2}\right)\) by two different process. The quantity which will remain same is
1 \(\Delta W\)
2 \(\Delta Q\)
3 \(\Delta Q-\Delta W\)
4 \(\Delta Q+\Delta W\)
Explanation:
Change in internal energy does not depend upon path so \(\Delta U=\Delta Q-\Delta W\) remains constant.
PHXI12:THERMODYNAMICS
371229
In thermodynamic process, \(200\,Joules\) of heat is given to a gas and \(100\,Joules\) of work is also done on it. The change in internal energy of the gas is
1 \(100\;J\)
2 \(300\;J\)
3 \(419\;J\)
4 \(24\;J\)
Explanation:
\(\Delta Q = \Delta U + \Delta W;\Delta Q = 200\,J\,{\mkern 1mu} {\rm{and}}{\mkern 1mu} \,\Delta W = - 100\,J\) \( \Rightarrow \Delta U = \Delta Q - \Delta W = 2000 - (100) = 300\,J.\)
PHXI12:THERMODYNAMICS
371230
An electric heater supplies heat to a system at a rate of \(120 \mathrm{~W}\). If system performs work at a rate of \(80\,J{s^{ - 1}}\)
1 \(30\,J{s^{ - 1}}\)
2 \(40\,J{s^{ - 1}}\)
3 \(50\,J{s^{ - 1}}\)
4 \(60\,J{s^{ - 1}}\)
Explanation:
According to first Law of thermodynamic \(\Delta Q=\Delta U+\Delta W\) \(\therefore \dfrac{\Delta Q}{\Delta t}=\dfrac{\Delta U}{\Delta t}+\dfrac{\Delta W}{\Delta t}\) Here, \(\frac{{\Delta Q}}{{\Delta t}} = 120W,\frac{{\Delta W}}{{\Delta t}} = 80J{s^{ - 1}}\) \(\therefore \frac{{\Delta U}}{{\Delta t}} = 120 - 80 = 40\,J{s^{ - 1}}\)
PHXI12:THERMODYNAMICS
371231
The change in internal energy of \(3\,mol\) of helium gas when its temperature is increased by \({2 K}\)
1 \(74.8\,cal\)
2 \(37.4\,cal\)
3 \(37.4\,J\)
4 \(74.8\,J\)
Explanation:
\({\Delta U=\mu C_{\nu} d T=\mu \times \dfrac{3}{2} R d T}\) \(=3 \times \dfrac{3}{2} \times 8.31 \times 2=74.8 {~J}\)
371228
A thermodynamical system is changed from state \(\left(p_{1} V_{1}\right)\) to \(\left(p_{2} V_{2}\right)\) by two different process. The quantity which will remain same is
1 \(\Delta W\)
2 \(\Delta Q\)
3 \(\Delta Q-\Delta W\)
4 \(\Delta Q+\Delta W\)
Explanation:
Change in internal energy does not depend upon path so \(\Delta U=\Delta Q-\Delta W\) remains constant.
PHXI12:THERMODYNAMICS
371229
In thermodynamic process, \(200\,Joules\) of heat is given to a gas and \(100\,Joules\) of work is also done on it. The change in internal energy of the gas is
1 \(100\;J\)
2 \(300\;J\)
3 \(419\;J\)
4 \(24\;J\)
Explanation:
\(\Delta Q = \Delta U + \Delta W;\Delta Q = 200\,J\,{\mkern 1mu} {\rm{and}}{\mkern 1mu} \,\Delta W = - 100\,J\) \( \Rightarrow \Delta U = \Delta Q - \Delta W = 2000 - (100) = 300\,J.\)
PHXI12:THERMODYNAMICS
371230
An electric heater supplies heat to a system at a rate of \(120 \mathrm{~W}\). If system performs work at a rate of \(80\,J{s^{ - 1}}\)
1 \(30\,J{s^{ - 1}}\)
2 \(40\,J{s^{ - 1}}\)
3 \(50\,J{s^{ - 1}}\)
4 \(60\,J{s^{ - 1}}\)
Explanation:
According to first Law of thermodynamic \(\Delta Q=\Delta U+\Delta W\) \(\therefore \dfrac{\Delta Q}{\Delta t}=\dfrac{\Delta U}{\Delta t}+\dfrac{\Delta W}{\Delta t}\) Here, \(\frac{{\Delta Q}}{{\Delta t}} = 120W,\frac{{\Delta W}}{{\Delta t}} = 80J{s^{ - 1}}\) \(\therefore \frac{{\Delta U}}{{\Delta t}} = 120 - 80 = 40\,J{s^{ - 1}}\)
PHXI12:THERMODYNAMICS
371231
The change in internal energy of \(3\,mol\) of helium gas when its temperature is increased by \({2 K}\)
1 \(74.8\,cal\)
2 \(37.4\,cal\)
3 \(37.4\,J\)
4 \(74.8\,J\)
Explanation:
\({\Delta U=\mu C_{\nu} d T=\mu \times \dfrac{3}{2} R d T}\) \(=3 \times \dfrac{3}{2} \times 8.31 \times 2=74.8 {~J}\)
371228
A thermodynamical system is changed from state \(\left(p_{1} V_{1}\right)\) to \(\left(p_{2} V_{2}\right)\) by two different process. The quantity which will remain same is
1 \(\Delta W\)
2 \(\Delta Q\)
3 \(\Delta Q-\Delta W\)
4 \(\Delta Q+\Delta W\)
Explanation:
Change in internal energy does not depend upon path so \(\Delta U=\Delta Q-\Delta W\) remains constant.
PHXI12:THERMODYNAMICS
371229
In thermodynamic process, \(200\,Joules\) of heat is given to a gas and \(100\,Joules\) of work is also done on it. The change in internal energy of the gas is
1 \(100\;J\)
2 \(300\;J\)
3 \(419\;J\)
4 \(24\;J\)
Explanation:
\(\Delta Q = \Delta U + \Delta W;\Delta Q = 200\,J\,{\mkern 1mu} {\rm{and}}{\mkern 1mu} \,\Delta W = - 100\,J\) \( \Rightarrow \Delta U = \Delta Q - \Delta W = 2000 - (100) = 300\,J.\)
PHXI12:THERMODYNAMICS
371230
An electric heater supplies heat to a system at a rate of \(120 \mathrm{~W}\). If system performs work at a rate of \(80\,J{s^{ - 1}}\)
1 \(30\,J{s^{ - 1}}\)
2 \(40\,J{s^{ - 1}}\)
3 \(50\,J{s^{ - 1}}\)
4 \(60\,J{s^{ - 1}}\)
Explanation:
According to first Law of thermodynamic \(\Delta Q=\Delta U+\Delta W\) \(\therefore \dfrac{\Delta Q}{\Delta t}=\dfrac{\Delta U}{\Delta t}+\dfrac{\Delta W}{\Delta t}\) Here, \(\frac{{\Delta Q}}{{\Delta t}} = 120W,\frac{{\Delta W}}{{\Delta t}} = 80J{s^{ - 1}}\) \(\therefore \frac{{\Delta U}}{{\Delta t}} = 120 - 80 = 40\,J{s^{ - 1}}\)
PHXI12:THERMODYNAMICS
371231
The change in internal energy of \(3\,mol\) of helium gas when its temperature is increased by \({2 K}\)
1 \(74.8\,cal\)
2 \(37.4\,cal\)
3 \(37.4\,J\)
4 \(74.8\,J\)
Explanation:
\({\Delta U=\mu C_{\nu} d T=\mu \times \dfrac{3}{2} R d T}\) \(=3 \times \dfrac{3}{2} \times 8.31 \times 2=74.8 {~J}\)
371228
A thermodynamical system is changed from state \(\left(p_{1} V_{1}\right)\) to \(\left(p_{2} V_{2}\right)\) by two different process. The quantity which will remain same is
1 \(\Delta W\)
2 \(\Delta Q\)
3 \(\Delta Q-\Delta W\)
4 \(\Delta Q+\Delta W\)
Explanation:
Change in internal energy does not depend upon path so \(\Delta U=\Delta Q-\Delta W\) remains constant.
PHXI12:THERMODYNAMICS
371229
In thermodynamic process, \(200\,Joules\) of heat is given to a gas and \(100\,Joules\) of work is also done on it. The change in internal energy of the gas is
1 \(100\;J\)
2 \(300\;J\)
3 \(419\;J\)
4 \(24\;J\)
Explanation:
\(\Delta Q = \Delta U + \Delta W;\Delta Q = 200\,J\,{\mkern 1mu} {\rm{and}}{\mkern 1mu} \,\Delta W = - 100\,J\) \( \Rightarrow \Delta U = \Delta Q - \Delta W = 2000 - (100) = 300\,J.\)
PHXI12:THERMODYNAMICS
371230
An electric heater supplies heat to a system at a rate of \(120 \mathrm{~W}\). If system performs work at a rate of \(80\,J{s^{ - 1}}\)
1 \(30\,J{s^{ - 1}}\)
2 \(40\,J{s^{ - 1}}\)
3 \(50\,J{s^{ - 1}}\)
4 \(60\,J{s^{ - 1}}\)
Explanation:
According to first Law of thermodynamic \(\Delta Q=\Delta U+\Delta W\) \(\therefore \dfrac{\Delta Q}{\Delta t}=\dfrac{\Delta U}{\Delta t}+\dfrac{\Delta W}{\Delta t}\) Here, \(\frac{{\Delta Q}}{{\Delta t}} = 120W,\frac{{\Delta W}}{{\Delta t}} = 80J{s^{ - 1}}\) \(\therefore \frac{{\Delta U}}{{\Delta t}} = 120 - 80 = 40\,J{s^{ - 1}}\)
PHXI12:THERMODYNAMICS
371231
The change in internal energy of \(3\,mol\) of helium gas when its temperature is increased by \({2 K}\)
1 \(74.8\,cal\)
2 \(37.4\,cal\)
3 \(37.4\,J\)
4 \(74.8\,J\)
Explanation:
\({\Delta U=\mu C_{\nu} d T=\mu \times \dfrac{3}{2} R d T}\) \(=3 \times \dfrac{3}{2} \times 8.31 \times 2=74.8 {~J}\)
