148500
For an adiabatic process, the relation between $V$ and $T$ is given by
1 $\mathrm{TV}^{\gamma}=$ constant
2 $\mathrm{T}^{\gamma} \mathrm{V}=$ constant
3 $\mathrm{TV}^{1-\gamma}=$ constant
4 $\mathrm{TV}^{\gamma-1}=$ constant
Explanation:
D For an adiabatic process $\mathrm{PV}^{\gamma}=\text { Constants }$ We know that, $\mathrm{PV}=\mathrm{RT}$ ( For one mole of gas) $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}}$ Substituting the value of $\mathrm{P}$ in equation (i), $\mathrm{PV}^{\gamma}=\mathrm{C}$ $\frac{\mathrm{RT}}{\mathrm{V}} \cdot \mathrm{V}^{\gamma}=\mathrm{C}$ $\mathrm{TV}^{\gamma-1}=\frac{\mathrm{C}}{\mathrm{R}}$ $\mathrm{TV}^{\gamma-1}=$ Constant
AP EAMCET(Medical)-2007
Thermodynamics
148501
One mole of an ideal gas with $\gamma=1.4$ is adiabatically compressed so that, its temperature rises from $27^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$. The change in the internal energy of gas is: $(\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mol}$ - K)
1 $-166 \mathrm{~J}$
2 $166 \mathrm{~J}$
3 $-168 \mathrm{~J}$
4 $168 \mathrm{~J}$
Explanation:
B Given that, $\gamma=1.4$ $\Delta \mathrm{T}=\left(35^{\circ}-27^{\circ}\right)=8$ $\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ We know that, The change in internal energy, $\Delta \mathrm{U} =\frac{\mathrm{n} \cdot \mathrm{R} \cdot \Delta \mathrm{t}}{\gamma-1}$ $\Delta \mathrm{U} =\frac{1 \times 8.3(8)}{1.4-1}$ $\Delta \mathrm{U} =166 \mathrm{~J}$
AP EAMCET(Medical)-2001
Thermodynamics
148502
An ideal gas at a pressure of $1 \mathrm{~atm}$ and temperature of $27^{\circ} \mathrm{C}$ is compressed adiabatically until its pressure becomes 8 times the initial pressure, then final temperature is $(\gamma$ $=3 / 2$ )
1 $627^{\circ} \mathrm{C}$
2 $527^{\circ} \mathrm{C}$
3 $427^{\circ} \mathrm{C}$
4 $327^{\circ} \mathrm{C}$
Explanation:
D Given that, $\mathrm{P}_{1}=1 \mathrm{~atm}, \mathrm{~T}_{2}=27^{\circ} \mathrm{C}=27+273$ $=300 \mathrm{~K}$ $\mathrm{P}_{2}=8 \mathrm{P}_{1}=8 \mathrm{~atm}, \gamma=\frac{3}{2}=1.5$ In an adiabatic process, $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{\gamma}=\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\left(\frac{\gamma-1}{\gamma}\right)}$ $\mathrm{T}_{2}=\mathrm{T}_{1} \times\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\left(\frac{\gamma-1}{\gamma}\right)}$ $=300 \times\left(\frac{8}{1}\right)^{\left(\frac{1.5-1}{1.5}\right)}$ $=300 \times\left(\frac{8}{1}\right)^{0.5 / 1.5}$ $\mathrm{~T}_{2}=600 \mathrm{~K}$ $\mathrm{~T}_{2}=(600-273)$ $=327^{\circ} \mathrm{C}$
EAMCET-2000
Thermodynamics
148503
In the adiabatic compression, the decrease in volume is associated with
1 increase in temperature and decrease in pressure
2 decrease in temperature and increase in pressure
3 decrease in temperature and decrease in pressure
4 increase in temperature and increase in pressure
Explanation:
D According to first law of thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dQ}=0$ from an adiabatic process $\mathrm{dU}=-\mathrm{dW}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)^{\frac{\gamma-1}{\gamma}}=\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\gamma-1}$ This relation suggest that an decrease in volume is associated with increase in temperature and pressure.
148500
For an adiabatic process, the relation between $V$ and $T$ is given by
1 $\mathrm{TV}^{\gamma}=$ constant
2 $\mathrm{T}^{\gamma} \mathrm{V}=$ constant
3 $\mathrm{TV}^{1-\gamma}=$ constant
4 $\mathrm{TV}^{\gamma-1}=$ constant
Explanation:
D For an adiabatic process $\mathrm{PV}^{\gamma}=\text { Constants }$ We know that, $\mathrm{PV}=\mathrm{RT}$ ( For one mole of gas) $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}}$ Substituting the value of $\mathrm{P}$ in equation (i), $\mathrm{PV}^{\gamma}=\mathrm{C}$ $\frac{\mathrm{RT}}{\mathrm{V}} \cdot \mathrm{V}^{\gamma}=\mathrm{C}$ $\mathrm{TV}^{\gamma-1}=\frac{\mathrm{C}}{\mathrm{R}}$ $\mathrm{TV}^{\gamma-1}=$ Constant
AP EAMCET(Medical)-2007
Thermodynamics
148501
One mole of an ideal gas with $\gamma=1.4$ is adiabatically compressed so that, its temperature rises from $27^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$. The change in the internal energy of gas is: $(\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mol}$ - K)
1 $-166 \mathrm{~J}$
2 $166 \mathrm{~J}$
3 $-168 \mathrm{~J}$
4 $168 \mathrm{~J}$
Explanation:
B Given that, $\gamma=1.4$ $\Delta \mathrm{T}=\left(35^{\circ}-27^{\circ}\right)=8$ $\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ We know that, The change in internal energy, $\Delta \mathrm{U} =\frac{\mathrm{n} \cdot \mathrm{R} \cdot \Delta \mathrm{t}}{\gamma-1}$ $\Delta \mathrm{U} =\frac{1 \times 8.3(8)}{1.4-1}$ $\Delta \mathrm{U} =166 \mathrm{~J}$
AP EAMCET(Medical)-2001
Thermodynamics
148502
An ideal gas at a pressure of $1 \mathrm{~atm}$ and temperature of $27^{\circ} \mathrm{C}$ is compressed adiabatically until its pressure becomes 8 times the initial pressure, then final temperature is $(\gamma$ $=3 / 2$ )
1 $627^{\circ} \mathrm{C}$
2 $527^{\circ} \mathrm{C}$
3 $427^{\circ} \mathrm{C}$
4 $327^{\circ} \mathrm{C}$
Explanation:
D Given that, $\mathrm{P}_{1}=1 \mathrm{~atm}, \mathrm{~T}_{2}=27^{\circ} \mathrm{C}=27+273$ $=300 \mathrm{~K}$ $\mathrm{P}_{2}=8 \mathrm{P}_{1}=8 \mathrm{~atm}, \gamma=\frac{3}{2}=1.5$ In an adiabatic process, $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{\gamma}=\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\left(\frac{\gamma-1}{\gamma}\right)}$ $\mathrm{T}_{2}=\mathrm{T}_{1} \times\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\left(\frac{\gamma-1}{\gamma}\right)}$ $=300 \times\left(\frac{8}{1}\right)^{\left(\frac{1.5-1}{1.5}\right)}$ $=300 \times\left(\frac{8}{1}\right)^{0.5 / 1.5}$ $\mathrm{~T}_{2}=600 \mathrm{~K}$ $\mathrm{~T}_{2}=(600-273)$ $=327^{\circ} \mathrm{C}$
EAMCET-2000
Thermodynamics
148503
In the adiabatic compression, the decrease in volume is associated with
1 increase in temperature and decrease in pressure
2 decrease in temperature and increase in pressure
3 decrease in temperature and decrease in pressure
4 increase in temperature and increase in pressure
Explanation:
D According to first law of thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dQ}=0$ from an adiabatic process $\mathrm{dU}=-\mathrm{dW}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)^{\frac{\gamma-1}{\gamma}}=\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\gamma-1}$ This relation suggest that an decrease in volume is associated with increase in temperature and pressure.
148500
For an adiabatic process, the relation between $V$ and $T$ is given by
1 $\mathrm{TV}^{\gamma}=$ constant
2 $\mathrm{T}^{\gamma} \mathrm{V}=$ constant
3 $\mathrm{TV}^{1-\gamma}=$ constant
4 $\mathrm{TV}^{\gamma-1}=$ constant
Explanation:
D For an adiabatic process $\mathrm{PV}^{\gamma}=\text { Constants }$ We know that, $\mathrm{PV}=\mathrm{RT}$ ( For one mole of gas) $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}}$ Substituting the value of $\mathrm{P}$ in equation (i), $\mathrm{PV}^{\gamma}=\mathrm{C}$ $\frac{\mathrm{RT}}{\mathrm{V}} \cdot \mathrm{V}^{\gamma}=\mathrm{C}$ $\mathrm{TV}^{\gamma-1}=\frac{\mathrm{C}}{\mathrm{R}}$ $\mathrm{TV}^{\gamma-1}=$ Constant
AP EAMCET(Medical)-2007
Thermodynamics
148501
One mole of an ideal gas with $\gamma=1.4$ is adiabatically compressed so that, its temperature rises from $27^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$. The change in the internal energy of gas is: $(\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mol}$ - K)
1 $-166 \mathrm{~J}$
2 $166 \mathrm{~J}$
3 $-168 \mathrm{~J}$
4 $168 \mathrm{~J}$
Explanation:
B Given that, $\gamma=1.4$ $\Delta \mathrm{T}=\left(35^{\circ}-27^{\circ}\right)=8$ $\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ We know that, The change in internal energy, $\Delta \mathrm{U} =\frac{\mathrm{n} \cdot \mathrm{R} \cdot \Delta \mathrm{t}}{\gamma-1}$ $\Delta \mathrm{U} =\frac{1 \times 8.3(8)}{1.4-1}$ $\Delta \mathrm{U} =166 \mathrm{~J}$
AP EAMCET(Medical)-2001
Thermodynamics
148502
An ideal gas at a pressure of $1 \mathrm{~atm}$ and temperature of $27^{\circ} \mathrm{C}$ is compressed adiabatically until its pressure becomes 8 times the initial pressure, then final temperature is $(\gamma$ $=3 / 2$ )
1 $627^{\circ} \mathrm{C}$
2 $527^{\circ} \mathrm{C}$
3 $427^{\circ} \mathrm{C}$
4 $327^{\circ} \mathrm{C}$
Explanation:
D Given that, $\mathrm{P}_{1}=1 \mathrm{~atm}, \mathrm{~T}_{2}=27^{\circ} \mathrm{C}=27+273$ $=300 \mathrm{~K}$ $\mathrm{P}_{2}=8 \mathrm{P}_{1}=8 \mathrm{~atm}, \gamma=\frac{3}{2}=1.5$ In an adiabatic process, $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{\gamma}=\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\left(\frac{\gamma-1}{\gamma}\right)}$ $\mathrm{T}_{2}=\mathrm{T}_{1} \times\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\left(\frac{\gamma-1}{\gamma}\right)}$ $=300 \times\left(\frac{8}{1}\right)^{\left(\frac{1.5-1}{1.5}\right)}$ $=300 \times\left(\frac{8}{1}\right)^{0.5 / 1.5}$ $\mathrm{~T}_{2}=600 \mathrm{~K}$ $\mathrm{~T}_{2}=(600-273)$ $=327^{\circ} \mathrm{C}$
EAMCET-2000
Thermodynamics
148503
In the adiabatic compression, the decrease in volume is associated with
1 increase in temperature and decrease in pressure
2 decrease in temperature and increase in pressure
3 decrease in temperature and decrease in pressure
4 increase in temperature and increase in pressure
Explanation:
D According to first law of thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dQ}=0$ from an adiabatic process $\mathrm{dU}=-\mathrm{dW}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)^{\frac{\gamma-1}{\gamma}}=\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\gamma-1}$ This relation suggest that an decrease in volume is associated with increase in temperature and pressure.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
148500
For an adiabatic process, the relation between $V$ and $T$ is given by
1 $\mathrm{TV}^{\gamma}=$ constant
2 $\mathrm{T}^{\gamma} \mathrm{V}=$ constant
3 $\mathrm{TV}^{1-\gamma}=$ constant
4 $\mathrm{TV}^{\gamma-1}=$ constant
Explanation:
D For an adiabatic process $\mathrm{PV}^{\gamma}=\text { Constants }$ We know that, $\mathrm{PV}=\mathrm{RT}$ ( For one mole of gas) $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}}$ Substituting the value of $\mathrm{P}$ in equation (i), $\mathrm{PV}^{\gamma}=\mathrm{C}$ $\frac{\mathrm{RT}}{\mathrm{V}} \cdot \mathrm{V}^{\gamma}=\mathrm{C}$ $\mathrm{TV}^{\gamma-1}=\frac{\mathrm{C}}{\mathrm{R}}$ $\mathrm{TV}^{\gamma-1}=$ Constant
AP EAMCET(Medical)-2007
Thermodynamics
148501
One mole of an ideal gas with $\gamma=1.4$ is adiabatically compressed so that, its temperature rises from $27^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$. The change in the internal energy of gas is: $(\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mol}$ - K)
1 $-166 \mathrm{~J}$
2 $166 \mathrm{~J}$
3 $-168 \mathrm{~J}$
4 $168 \mathrm{~J}$
Explanation:
B Given that, $\gamma=1.4$ $\Delta \mathrm{T}=\left(35^{\circ}-27^{\circ}\right)=8$ $\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ We know that, The change in internal energy, $\Delta \mathrm{U} =\frac{\mathrm{n} \cdot \mathrm{R} \cdot \Delta \mathrm{t}}{\gamma-1}$ $\Delta \mathrm{U} =\frac{1 \times 8.3(8)}{1.4-1}$ $\Delta \mathrm{U} =166 \mathrm{~J}$
AP EAMCET(Medical)-2001
Thermodynamics
148502
An ideal gas at a pressure of $1 \mathrm{~atm}$ and temperature of $27^{\circ} \mathrm{C}$ is compressed adiabatically until its pressure becomes 8 times the initial pressure, then final temperature is $(\gamma$ $=3 / 2$ )
1 $627^{\circ} \mathrm{C}$
2 $527^{\circ} \mathrm{C}$
3 $427^{\circ} \mathrm{C}$
4 $327^{\circ} \mathrm{C}$
Explanation:
D Given that, $\mathrm{P}_{1}=1 \mathrm{~atm}, \mathrm{~T}_{2}=27^{\circ} \mathrm{C}=27+273$ $=300 \mathrm{~K}$ $\mathrm{P}_{2}=8 \mathrm{P}_{1}=8 \mathrm{~atm}, \gamma=\frac{3}{2}=1.5$ In an adiabatic process, $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{\gamma}=\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\left(\frac{\gamma-1}{\gamma}\right)}$ $\mathrm{T}_{2}=\mathrm{T}_{1} \times\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right)^{\left(\frac{\gamma-1}{\gamma}\right)}$ $=300 \times\left(\frac{8}{1}\right)^{\left(\frac{1.5-1}{1.5}\right)}$ $=300 \times\left(\frac{8}{1}\right)^{0.5 / 1.5}$ $\mathrm{~T}_{2}=600 \mathrm{~K}$ $\mathrm{~T}_{2}=(600-273)$ $=327^{\circ} \mathrm{C}$
EAMCET-2000
Thermodynamics
148503
In the adiabatic compression, the decrease in volume is associated with
1 increase in temperature and decrease in pressure
2 decrease in temperature and increase in pressure
3 decrease in temperature and decrease in pressure
4 increase in temperature and increase in pressure
Explanation:
D According to first law of thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dQ}=0$ from an adiabatic process $\mathrm{dU}=-\mathrm{dW}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)^{\frac{\gamma-1}{\gamma}}=\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\gamma-1}$ This relation suggest that an decrease in volume is associated with increase in temperature and pressure.
