148115
Two moles of a monoatomic ideal gas is confined in a container and is heated such that its temperature increases by $10^{\circ} \mathrm{C}$. The approximate change in its internal energy is $(\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mole}-\mathrm{K})$
148117
The latent heat of vaporization of water is 2240 $\mathrm{J} / \mathrm{g}$. If the work done in the process of vaporization of $1 \mathrm{~g}$ is $168 \mathrm{~J}$, then increase in internal energy is
1 $1940 \mathrm{~J}$
2 $2072 \mathrm{~J}$
3 $2240 \mathrm{~J}$
4 $2408 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{L}_{\mathrm{V}}=2240 \mathrm{~J} / \mathrm{g}$, mass, $\mathrm{m}=1 \mathrm{~g}$, $\Delta \mathrm{W}=168 \mathrm{~J}$ $\because \quad \Delta \mathrm{Q}=\mathrm{mL}_{\mathrm{V}}$ $=1 \times 2240$ $=2240 \mathrm{~J}$ According to the first law of thermodynamics- $\Delta \mathrm{Q} =\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{U} =\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=2240-168=2072 \mathrm{~J}$
Punjab PET 1998
Thermodynamics
148118
A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process? $(\mathrm{J}=\mathbf{4 . 1 8} \mathrm{Joules} / \mathrm{cal})$
1 654 Joule
2 156.5 Joule
3 -300 Joule
4 -528.2 Joule
Explanation:
A Given, $\Delta \mathrm{Q}=300 \mathrm{cal}=300 \times 4.18 \mathrm{~J}=1254$ $\mathrm{J}, \Delta \mathrm{W}=600 \mathrm{~J}$ According to first law of thermodynamics- $\Delta \mathrm{Q} =\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{U} =\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=1254-600$ $=654 \mathrm{~J}$
148115
Two moles of a monoatomic ideal gas is confined in a container and is heated such that its temperature increases by $10^{\circ} \mathrm{C}$. The approximate change in its internal energy is $(\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mole}-\mathrm{K})$
148117
The latent heat of vaporization of water is 2240 $\mathrm{J} / \mathrm{g}$. If the work done in the process of vaporization of $1 \mathrm{~g}$ is $168 \mathrm{~J}$, then increase in internal energy is
1 $1940 \mathrm{~J}$
2 $2072 \mathrm{~J}$
3 $2240 \mathrm{~J}$
4 $2408 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{L}_{\mathrm{V}}=2240 \mathrm{~J} / \mathrm{g}$, mass, $\mathrm{m}=1 \mathrm{~g}$, $\Delta \mathrm{W}=168 \mathrm{~J}$ $\because \quad \Delta \mathrm{Q}=\mathrm{mL}_{\mathrm{V}}$ $=1 \times 2240$ $=2240 \mathrm{~J}$ According to the first law of thermodynamics- $\Delta \mathrm{Q} =\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{U} =\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=2240-168=2072 \mathrm{~J}$
Punjab PET 1998
Thermodynamics
148118
A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process? $(\mathrm{J}=\mathbf{4 . 1 8} \mathrm{Joules} / \mathrm{cal})$
1 654 Joule
2 156.5 Joule
3 -300 Joule
4 -528.2 Joule
Explanation:
A Given, $\Delta \mathrm{Q}=300 \mathrm{cal}=300 \times 4.18 \mathrm{~J}=1254$ $\mathrm{J}, \Delta \mathrm{W}=600 \mathrm{~J}$ According to first law of thermodynamics- $\Delta \mathrm{Q} =\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{U} =\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=1254-600$ $=654 \mathrm{~J}$
148115
Two moles of a monoatomic ideal gas is confined in a container and is heated such that its temperature increases by $10^{\circ} \mathrm{C}$. The approximate change in its internal energy is $(\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mole}-\mathrm{K})$
148117
The latent heat of vaporization of water is 2240 $\mathrm{J} / \mathrm{g}$. If the work done in the process of vaporization of $1 \mathrm{~g}$ is $168 \mathrm{~J}$, then increase in internal energy is
1 $1940 \mathrm{~J}$
2 $2072 \mathrm{~J}$
3 $2240 \mathrm{~J}$
4 $2408 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{L}_{\mathrm{V}}=2240 \mathrm{~J} / \mathrm{g}$, mass, $\mathrm{m}=1 \mathrm{~g}$, $\Delta \mathrm{W}=168 \mathrm{~J}$ $\because \quad \Delta \mathrm{Q}=\mathrm{mL}_{\mathrm{V}}$ $=1 \times 2240$ $=2240 \mathrm{~J}$ According to the first law of thermodynamics- $\Delta \mathrm{Q} =\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{U} =\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=2240-168=2072 \mathrm{~J}$
Punjab PET 1998
Thermodynamics
148118
A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process? $(\mathrm{J}=\mathbf{4 . 1 8} \mathrm{Joules} / \mathrm{cal})$
1 654 Joule
2 156.5 Joule
3 -300 Joule
4 -528.2 Joule
Explanation:
A Given, $\Delta \mathrm{Q}=300 \mathrm{cal}=300 \times 4.18 \mathrm{~J}=1254$ $\mathrm{J}, \Delta \mathrm{W}=600 \mathrm{~J}$ According to first law of thermodynamics- $\Delta \mathrm{Q} =\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{U} =\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=1254-600$ $=654 \mathrm{~J}$
148115
Two moles of a monoatomic ideal gas is confined in a container and is heated such that its temperature increases by $10^{\circ} \mathrm{C}$. The approximate change in its internal energy is $(\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mole}-\mathrm{K})$
148117
The latent heat of vaporization of water is 2240 $\mathrm{J} / \mathrm{g}$. If the work done in the process of vaporization of $1 \mathrm{~g}$ is $168 \mathrm{~J}$, then increase in internal energy is
1 $1940 \mathrm{~J}$
2 $2072 \mathrm{~J}$
3 $2240 \mathrm{~J}$
4 $2408 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{L}_{\mathrm{V}}=2240 \mathrm{~J} / \mathrm{g}$, mass, $\mathrm{m}=1 \mathrm{~g}$, $\Delta \mathrm{W}=168 \mathrm{~J}$ $\because \quad \Delta \mathrm{Q}=\mathrm{mL}_{\mathrm{V}}$ $=1 \times 2240$ $=2240 \mathrm{~J}$ According to the first law of thermodynamics- $\Delta \mathrm{Q} =\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{U} =\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=2240-168=2072 \mathrm{~J}$
Punjab PET 1998
Thermodynamics
148118
A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process? $(\mathrm{J}=\mathbf{4 . 1 8} \mathrm{Joules} / \mathrm{cal})$
1 654 Joule
2 156.5 Joule
3 -300 Joule
4 -528.2 Joule
Explanation:
A Given, $\Delta \mathrm{Q}=300 \mathrm{cal}=300 \times 4.18 \mathrm{~J}=1254$ $\mathrm{J}, \Delta \mathrm{W}=600 \mathrm{~J}$ According to first law of thermodynamics- $\Delta \mathrm{Q} =\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{U} =\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=1254-600$ $=654 \mathrm{~J}$