148471
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of absolute temperature. The ratio $\frac{C_{p}}{C_{v}}$ for the gas is
1 $\frac{4}{3}$
2 2
3 $\frac{5}{3}$
4 $\frac{3}{2}$
Explanation:
D For an adiabatic process relation between pressure and temperature is given as- $\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{\frac{\gamma}{\gamma-1}}$ $\mathrm{P} \propto(\mathrm{T})^{\frac{\gamma}{\gamma-1}}$ And according to question $\mathrm{P} \propto \mathrm{T}^{3}$. Comparing equation (i) from equation (ii) we, get- $\frac{\gamma}{\gamma-1}=3$ $\gamma=3 \gamma-3$ $2 \gamma=3$ $\gamma=\frac{3}{2}$ $\therefore$ We know that $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{3}{2}$
TS EAMCET 09.05.2019
Thermodynamics
148472
The work of $146 \mathrm{~kJ}$ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by $7^{\circ} \mathrm{C}$. The gas is $(R=8.3 \mathrm{~J}$ mol-1 $\mathbf{K}^{-1}$ )
1 diatomic
2 triatomic
3 a mixture of monoatomic and diatomic
4 monoatomic
Explanation:
A $\mathrm{W}=-146 \mathrm{~kJ}$ $\mathrm{n}=1000 \text { moles. }$ $\mathrm{T}_{2}-\mathrm{T}_{1}=7{ }^{0} \mathrm{C}$ We know that, $\mathrm{W}=\frac{\mathrm{nR}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\gamma-1}$ $146000= \frac{1000 \times 8.3 \times 7}{\gamma-1}$ $\gamma-1=.4$ $\gamma=1.4$ Hence, the gas is diatomic gas.
UP CPMT-2008
Thermodynamics
148473
A perfect gas is found to obey the relation $\mathbf{P V}^{3 / 2}=$ constant during an adiabatic process. If such a gas initially at a temperature $T$, is compressed to half of its initial volume, then its final temperature will be:
148471
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of absolute temperature. The ratio $\frac{C_{p}}{C_{v}}$ for the gas is
1 $\frac{4}{3}$
2 2
3 $\frac{5}{3}$
4 $\frac{3}{2}$
Explanation:
D For an adiabatic process relation between pressure and temperature is given as- $\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{\frac{\gamma}{\gamma-1}}$ $\mathrm{P} \propto(\mathrm{T})^{\frac{\gamma}{\gamma-1}}$ And according to question $\mathrm{P} \propto \mathrm{T}^{3}$. Comparing equation (i) from equation (ii) we, get- $\frac{\gamma}{\gamma-1}=3$ $\gamma=3 \gamma-3$ $2 \gamma=3$ $\gamma=\frac{3}{2}$ $\therefore$ We know that $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{3}{2}$
TS EAMCET 09.05.2019
Thermodynamics
148472
The work of $146 \mathrm{~kJ}$ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by $7^{\circ} \mathrm{C}$. The gas is $(R=8.3 \mathrm{~J}$ mol-1 $\mathbf{K}^{-1}$ )
1 diatomic
2 triatomic
3 a mixture of monoatomic and diatomic
4 monoatomic
Explanation:
A $\mathrm{W}=-146 \mathrm{~kJ}$ $\mathrm{n}=1000 \text { moles. }$ $\mathrm{T}_{2}-\mathrm{T}_{1}=7{ }^{0} \mathrm{C}$ We know that, $\mathrm{W}=\frac{\mathrm{nR}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\gamma-1}$ $146000= \frac{1000 \times 8.3 \times 7}{\gamma-1}$ $\gamma-1=.4$ $\gamma=1.4$ Hence, the gas is diatomic gas.
UP CPMT-2008
Thermodynamics
148473
A perfect gas is found to obey the relation $\mathbf{P V}^{3 / 2}=$ constant during an adiabatic process. If such a gas initially at a temperature $T$, is compressed to half of its initial volume, then its final temperature will be:
NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
148471
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of absolute temperature. The ratio $\frac{C_{p}}{C_{v}}$ for the gas is
1 $\frac{4}{3}$
2 2
3 $\frac{5}{3}$
4 $\frac{3}{2}$
Explanation:
D For an adiabatic process relation between pressure and temperature is given as- $\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{\frac{\gamma}{\gamma-1}}$ $\mathrm{P} \propto(\mathrm{T})^{\frac{\gamma}{\gamma-1}}$ And according to question $\mathrm{P} \propto \mathrm{T}^{3}$. Comparing equation (i) from equation (ii) we, get- $\frac{\gamma}{\gamma-1}=3$ $\gamma=3 \gamma-3$ $2 \gamma=3$ $\gamma=\frac{3}{2}$ $\therefore$ We know that $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{3}{2}$
TS EAMCET 09.05.2019
Thermodynamics
148472
The work of $146 \mathrm{~kJ}$ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by $7^{\circ} \mathrm{C}$. The gas is $(R=8.3 \mathrm{~J}$ mol-1 $\mathbf{K}^{-1}$ )
1 diatomic
2 triatomic
3 a mixture of monoatomic and diatomic
4 monoatomic
Explanation:
A $\mathrm{W}=-146 \mathrm{~kJ}$ $\mathrm{n}=1000 \text { moles. }$ $\mathrm{T}_{2}-\mathrm{T}_{1}=7{ }^{0} \mathrm{C}$ We know that, $\mathrm{W}=\frac{\mathrm{nR}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\gamma-1}$ $146000= \frac{1000 \times 8.3 \times 7}{\gamma-1}$ $\gamma-1=.4$ $\gamma=1.4$ Hence, the gas is diatomic gas.
UP CPMT-2008
Thermodynamics
148473
A perfect gas is found to obey the relation $\mathbf{P V}^{3 / 2}=$ constant during an adiabatic process. If such a gas initially at a temperature $T$, is compressed to half of its initial volume, then its final temperature will be:
148471
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of absolute temperature. The ratio $\frac{C_{p}}{C_{v}}$ for the gas is
1 $\frac{4}{3}$
2 2
3 $\frac{5}{3}$
4 $\frac{3}{2}$
Explanation:
D For an adiabatic process relation between pressure and temperature is given as- $\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{\frac{\gamma}{\gamma-1}}$ $\mathrm{P} \propto(\mathrm{T})^{\frac{\gamma}{\gamma-1}}$ And according to question $\mathrm{P} \propto \mathrm{T}^{3}$. Comparing equation (i) from equation (ii) we, get- $\frac{\gamma}{\gamma-1}=3$ $\gamma=3 \gamma-3$ $2 \gamma=3$ $\gamma=\frac{3}{2}$ $\therefore$ We know that $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{3}{2}$
TS EAMCET 09.05.2019
Thermodynamics
148472
The work of $146 \mathrm{~kJ}$ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by $7^{\circ} \mathrm{C}$. The gas is $(R=8.3 \mathrm{~J}$ mol-1 $\mathbf{K}^{-1}$ )
1 diatomic
2 triatomic
3 a mixture of monoatomic and diatomic
4 monoatomic
Explanation:
A $\mathrm{W}=-146 \mathrm{~kJ}$ $\mathrm{n}=1000 \text { moles. }$ $\mathrm{T}_{2}-\mathrm{T}_{1}=7{ }^{0} \mathrm{C}$ We know that, $\mathrm{W}=\frac{\mathrm{nR}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\gamma-1}$ $146000= \frac{1000 \times 8.3 \times 7}{\gamma-1}$ $\gamma-1=.4$ $\gamma=1.4$ Hence, the gas is diatomic gas.
UP CPMT-2008
Thermodynamics
148473
A perfect gas is found to obey the relation $\mathbf{P V}^{3 / 2}=$ constant during an adiabatic process. If such a gas initially at a temperature $T$, is compressed to half of its initial volume, then its final temperature will be: