In a given process on an ideal gas dw = 0 and dQ < 0. Then for the gas
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Explanation
dQ = du + dw
Physics MCQs for NEET — Practice Questions with Answers
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Wafer of volume 2 filter in a containes is heated with a coil of 1kw at $ 27^\circ C $ The lid of the containes is open and energy dissipates at the late of 160 J/S In how much time tempreture will rise from $ 27 ^\circ to 77 ^\circ $. Specific heat of wafers is 4.2 KJ /Kg
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Explanation
energy required in heating water = $ ms \triangle \theta $ from coil, $ { , energy avilable \over S } = Power of coil - Power lost $ For 840 J time required = 1S $ \therefore 4.2 \times 10 ^ {4} J = ? $ $ t = { 4.2 \times 10 ^ 5 \over 840 } 500 J = 8.33 min = 8 min 205 $
70 calorie of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from $ 30 ^\circ to 35 ^\circ $ .The amount of heat required to raise the temperature of the same gas through the same range at constant volume is............calorie.
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Explanation
$ C_p = { ( \triangle Q ) P \over r \triangle T }$ cv = cp - R = 7-2 = 5 cal /mol K
When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is..
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Explanation
$ { \triangle u \over \triangle Q } = { 1 \over \wp } = { 5 \over 7 } $ $ \triangle u = n C_v \triangle T $ $ \triangle Q = n C_p \triangle T $ $ { \triangle u \over \triangle Q } = { C_v \over C_p } = { 1 \over r } = { 5 \over 7 } $
An insulated containes containing monoatomic gas of molas mass Mo is moving with avelocity, V.If the container is suddenly stopped, find the change in temperature.
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Explanation
$ Decrese in K.E = increase \;in I.E { 1 \over 2} mv^2 = \gamma C_v \triangle T $
A Small spherical body of radius r is falling under gravity in a viscous medium. Due to friction the medium gets heated. How does the late of heating depend on radius of body when it attains terminal velocity!
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Explanation
$ Rate of heat produced = (Viscous force F) \times (Velocity V) $ $ { dQ \over dt} = 6 \pi nr v^2 \left[ { 2 \over g } { ( \rho - \rho_o)r^2 g \over n } \right] ^2 $
The first law of thermodynamics is concerned withthe conservation of
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Explanation
$ \triangle Q = \triangle U + \triangle W $
If heat givento a system is 6 k cal and work done is 6kj. The change in internal energy is ...........KJ.
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The internal energy change in a system that has absorbed 2 Kcal of heat and done 500J of work is
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Explanation
$ \triangle Q = \triangle U + \triangle W $
Which of the following is not a thermodynamical function.
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Explanation
Work done is not a thermodynamic function because it depends on the path taken, rather than being a state function like Enthalpy, Gibbs energy, and Internal energy, which depend only on the initial and final states of the system.
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