148408
A monoatomic gas of pressure ' $P$ ' having volume ' $V$ ' expands isothermally to a volume ' $2 \mathrm{~V}$ ' and then adiabatically to a volume ' $16 \mathrm{~V}$ '. The final pressure of the gas is (ratio of specific heats $=\frac{5}{3}$ )
C Work done $=\int$ P.dV _..... (i) $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{n}=1$ mole $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}}$ Work done $=\mathrm{RT} \int_{\mathrm{V}_{1}}^{\mathrm{V}_{2}} \frac{\mathrm{dV}}{\mathrm{V}}$ $=\mathrm{RT}\left[\ln \mathrm{V}_{\mathrm{V}_{1}}^{\mathrm{V}_{2}}\right.$ $=\mathrm{RT}\left(\ln \mathrm{V}_{2}-\ln \mathrm{V}_{1}\right)$ $=\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ $\{\ln \left.=\log _{\mathrm{e}}\right\}$ $\therefore$ Work done $=\mathrm{RT} \log _{\mathrm{e}}\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$
UP CPMT-2010
Thermodynamics
148415
In an isothermal process if heat is supplied to an ideal gas, then
1 The internal energy of the gas will decrease
2 The internal energy of the gas will increase
3 The gas will do positive work
4 The gas will do negative work
Explanation:
C In Isothermal process the temperature is constant. The internal energy is a state function dependent on temperature. Hence, the internal energy change is zero. For an ideal gas, in an isothermal Process - $\Delta \mathrm{U}=0=\mathrm{Q}-\mathrm{W}$ So, $\mathrm{W}=\mathrm{Q}$ Hence, gas will do positive work.
TS EAMCET (Medical) 09.08.2021
Thermodynamics
148416
The pressure of a gas kept in an isothermal container is $200 \mathrm{kPa}$. If half the gas is removed from it, the pressure will be:
1 $100 \mathrm{kPa}$
2 $200 \mathrm{kPa}$
3 $400 \mathrm{kPa}$
4 $800 \mathrm{kPa}$
Explanation:
A Let the number of mole $\mathrm{n}$ and then become $\mathrm{n} / 2$. $\mathrm{n}_{1}=\mathrm{n}, \mathrm{n}_{2}=\frac{\mathrm{n}}{2}$ and $\mathrm{P}_{1}=200 \mathrm{kPa}$ We know that, $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \mathrm{P} \propto \mathrm{n}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\frac{\mathrm{n}}{2}}{\mathrm{n}}=\frac{1}{2}$ $\mathrm{P}_{2}=\frac{1}{2} \mathrm{P}_{1}$ $=\frac{1}{2} \times 200$ $=100 \mathrm{kPa}$
148408
A monoatomic gas of pressure ' $P$ ' having volume ' $V$ ' expands isothermally to a volume ' $2 \mathrm{~V}$ ' and then adiabatically to a volume ' $16 \mathrm{~V}$ '. The final pressure of the gas is (ratio of specific heats $=\frac{5}{3}$ )
C Work done $=\int$ P.dV _..... (i) $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{n}=1$ mole $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}}$ Work done $=\mathrm{RT} \int_{\mathrm{V}_{1}}^{\mathrm{V}_{2}} \frac{\mathrm{dV}}{\mathrm{V}}$ $=\mathrm{RT}\left[\ln \mathrm{V}_{\mathrm{V}_{1}}^{\mathrm{V}_{2}}\right.$ $=\mathrm{RT}\left(\ln \mathrm{V}_{2}-\ln \mathrm{V}_{1}\right)$ $=\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ $\{\ln \left.=\log _{\mathrm{e}}\right\}$ $\therefore$ Work done $=\mathrm{RT} \log _{\mathrm{e}}\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$
UP CPMT-2010
Thermodynamics
148415
In an isothermal process if heat is supplied to an ideal gas, then
1 The internal energy of the gas will decrease
2 The internal energy of the gas will increase
3 The gas will do positive work
4 The gas will do negative work
Explanation:
C In Isothermal process the temperature is constant. The internal energy is a state function dependent on temperature. Hence, the internal energy change is zero. For an ideal gas, in an isothermal Process - $\Delta \mathrm{U}=0=\mathrm{Q}-\mathrm{W}$ So, $\mathrm{W}=\mathrm{Q}$ Hence, gas will do positive work.
TS EAMCET (Medical) 09.08.2021
Thermodynamics
148416
The pressure of a gas kept in an isothermal container is $200 \mathrm{kPa}$. If half the gas is removed from it, the pressure will be:
1 $100 \mathrm{kPa}$
2 $200 \mathrm{kPa}$
3 $400 \mathrm{kPa}$
4 $800 \mathrm{kPa}$
Explanation:
A Let the number of mole $\mathrm{n}$ and then become $\mathrm{n} / 2$. $\mathrm{n}_{1}=\mathrm{n}, \mathrm{n}_{2}=\frac{\mathrm{n}}{2}$ and $\mathrm{P}_{1}=200 \mathrm{kPa}$ We know that, $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \mathrm{P} \propto \mathrm{n}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\frac{\mathrm{n}}{2}}{\mathrm{n}}=\frac{1}{2}$ $\mathrm{P}_{2}=\frac{1}{2} \mathrm{P}_{1}$ $=\frac{1}{2} \times 200$ $=100 \mathrm{kPa}$
148408
A monoatomic gas of pressure ' $P$ ' having volume ' $V$ ' expands isothermally to a volume ' $2 \mathrm{~V}$ ' and then adiabatically to a volume ' $16 \mathrm{~V}$ '. The final pressure of the gas is (ratio of specific heats $=\frac{5}{3}$ )
C Work done $=\int$ P.dV _..... (i) $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{n}=1$ mole $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}}$ Work done $=\mathrm{RT} \int_{\mathrm{V}_{1}}^{\mathrm{V}_{2}} \frac{\mathrm{dV}}{\mathrm{V}}$ $=\mathrm{RT}\left[\ln \mathrm{V}_{\mathrm{V}_{1}}^{\mathrm{V}_{2}}\right.$ $=\mathrm{RT}\left(\ln \mathrm{V}_{2}-\ln \mathrm{V}_{1}\right)$ $=\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ $\{\ln \left.=\log _{\mathrm{e}}\right\}$ $\therefore$ Work done $=\mathrm{RT} \log _{\mathrm{e}}\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$
UP CPMT-2010
Thermodynamics
148415
In an isothermal process if heat is supplied to an ideal gas, then
1 The internal energy of the gas will decrease
2 The internal energy of the gas will increase
3 The gas will do positive work
4 The gas will do negative work
Explanation:
C In Isothermal process the temperature is constant. The internal energy is a state function dependent on temperature. Hence, the internal energy change is zero. For an ideal gas, in an isothermal Process - $\Delta \mathrm{U}=0=\mathrm{Q}-\mathrm{W}$ So, $\mathrm{W}=\mathrm{Q}$ Hence, gas will do positive work.
TS EAMCET (Medical) 09.08.2021
Thermodynamics
148416
The pressure of a gas kept in an isothermal container is $200 \mathrm{kPa}$. If half the gas is removed from it, the pressure will be:
1 $100 \mathrm{kPa}$
2 $200 \mathrm{kPa}$
3 $400 \mathrm{kPa}$
4 $800 \mathrm{kPa}$
Explanation:
A Let the number of mole $\mathrm{n}$ and then become $\mathrm{n} / 2$. $\mathrm{n}_{1}=\mathrm{n}, \mathrm{n}_{2}=\frac{\mathrm{n}}{2}$ and $\mathrm{P}_{1}=200 \mathrm{kPa}$ We know that, $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \mathrm{P} \propto \mathrm{n}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\frac{\mathrm{n}}{2}}{\mathrm{n}}=\frac{1}{2}$ $\mathrm{P}_{2}=\frac{1}{2} \mathrm{P}_{1}$ $=\frac{1}{2} \times 200$ $=100 \mathrm{kPa}$
148408
A monoatomic gas of pressure ' $P$ ' having volume ' $V$ ' expands isothermally to a volume ' $2 \mathrm{~V}$ ' and then adiabatically to a volume ' $16 \mathrm{~V}$ '. The final pressure of the gas is (ratio of specific heats $=\frac{5}{3}$ )
C Work done $=\int$ P.dV _..... (i) $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{n}=1$ mole $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}}$ Work done $=\mathrm{RT} \int_{\mathrm{V}_{1}}^{\mathrm{V}_{2}} \frac{\mathrm{dV}}{\mathrm{V}}$ $=\mathrm{RT}\left[\ln \mathrm{V}_{\mathrm{V}_{1}}^{\mathrm{V}_{2}}\right.$ $=\mathrm{RT}\left(\ln \mathrm{V}_{2}-\ln \mathrm{V}_{1}\right)$ $=\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ $\{\ln \left.=\log _{\mathrm{e}}\right\}$ $\therefore$ Work done $=\mathrm{RT} \log _{\mathrm{e}}\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$
UP CPMT-2010
Thermodynamics
148415
In an isothermal process if heat is supplied to an ideal gas, then
1 The internal energy of the gas will decrease
2 The internal energy of the gas will increase
3 The gas will do positive work
4 The gas will do negative work
Explanation:
C In Isothermal process the temperature is constant. The internal energy is a state function dependent on temperature. Hence, the internal energy change is zero. For an ideal gas, in an isothermal Process - $\Delta \mathrm{U}=0=\mathrm{Q}-\mathrm{W}$ So, $\mathrm{W}=\mathrm{Q}$ Hence, gas will do positive work.
TS EAMCET (Medical) 09.08.2021
Thermodynamics
148416
The pressure of a gas kept in an isothermal container is $200 \mathrm{kPa}$. If half the gas is removed from it, the pressure will be:
1 $100 \mathrm{kPa}$
2 $200 \mathrm{kPa}$
3 $400 \mathrm{kPa}$
4 $800 \mathrm{kPa}$
Explanation:
A Let the number of mole $\mathrm{n}$ and then become $\mathrm{n} / 2$. $\mathrm{n}_{1}=\mathrm{n}, \mathrm{n}_{2}=\frac{\mathrm{n}}{2}$ and $\mathrm{P}_{1}=200 \mathrm{kPa}$ We know that, $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \mathrm{P} \propto \mathrm{n}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\frac{\mathrm{n}}{2}}{\mathrm{n}}=\frac{1}{2}$ $\mathrm{P}_{2}=\frac{1}{2} \mathrm{P}_{1}$ $=\frac{1}{2} \times 200$ $=100 \mathrm{kPa}$
