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The Volume of an ideal gas is 1 liter column and its Pressure is equal to 72 cm of Hg. The Volume of gas is made $ 900 cm^3 $ by compressing it isothermally. The stress of the gas will be...................... Hg column.
$P_1V_2 = P_2V_2$ (isothermal process)
In adiabatic expansion
$ \triangle Q = \triangle u+ \triangle W $ $ 0 = \triangle u + \triangle W $
$ 1 mm^3 $ of a gas is compressed at 1 atmospheric pressure and temperature $ 27 ^\circ C to 627 ^\circ C $ What is the final pressure under adiabatic condition. r = 1.5
$ {T^{\gamma} \over P^{\gamma_1}} = constant [ adiabatic change ] $ $ \left( { P2 \over P1} \right) ^ {1/2 } =\left( {T_2 \over T_1} \right) ^ { 3/2} $ $ \left( { P2 \over 10^5} \right) ^{1/2} = \left({ 900 \over 300 } \right) ^ {3/2} $
A monoatomic gas for it $ \gamma = { 5 \over 3} $ is suddenly Compressed to $ { 1 \over 8} $ of its original volume adiabatically then the final Pressure of gas is................ times its intial Pressure
$ PV \wp$ = Constant (adiabatic compressed) $ \left( { P_2 \over P_1 } \right) = \left( { V_1 \over V_2 } \right)^8 $
The Pressure and density of a diatomic gas $ \gamma = { 7 \over 5} $ Change adiabatically from (P,d) to $(P^1,d^1)$ . $ { d' \over d } =32 $ then $ { p' \over p} $ should be
$ PV^{\gamma}$ = Constant (adiabatic process)
An ideal gas at $ 27 ^\circ C$ is Compressed adiabatically, to $ { 8 \over 27 } $ of its original Volume . If $ \nu = { 5 \over 3} $ , then the rise in temperaure is
$ { T_2 \over T_1} = \left({ V_1 \over V_2} \right)^{\gamma - 1 } $
A diatomic gas intially at $ 18 ^\circ C $ is Compressed adiabatically to one eight of its original volume. The temperature after Compression will be
$ TV^{\gamma - 1 } $ = constant $ T_2 = T_1 \left( { V_1 \over V_2 } \right) ^ { \gamma -1} $
Work done by 0.1 mole of a gas at $ 27 ^ \circ C $ to double its volume at constant Pressure is ........$ Cal. R = 2 Cal / mol ^\circ K $
$ { V_1 \over V_2 } = { T_1 \over T_2} $ $ \therefore { V \over 2V} = { 300 \over T_2 } $ $ \therefore T_2 = 600 K $ $ W = P \triangle V $
A gas expnds $ 0.25 m^3 $ at constant pressure $ 10 ^3 N /m^2 $ the work done is
$ W = P \triangle V = 10 ^3 \times 0.25 = 250 J $
In an isochoric Process $ T_1 = 27 ^ \circ C and T_2 = 127 ^ \circ C $ then ${ P_1 \over P_2} $ will be equal to
$ P \alpha T ( constant value ) $ $ { P_1 \over P_2} = { T_1 \over T_2} $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.