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JEE PHYSICS MCQ: THERMODYNAMICS

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One gram of water (volume =1 cm3=1 \mathrm{~cm}^3 ) becomes 1671 cm3\mathrm{cm}^3 of steam when boiled at a pressure of one atmosphere . Latent heat of vaporization at this pressure is 539 cal/gm539 \mathrm{~cal} / \mathrm{gm}. Compute the work done. [1 atm=1.013×105Nm−2\left[1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{Nm}^{-2}\right. ]
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Correct option: A
169
The correct answer is 169 because the work done is calculated using the formula W = P(V2 - V1) with the given pressure and change in volume.

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The work done in which of the following process is zero? Water is used to cool the radiators of engines in cars because A gas expand 0.25  m30.25\,\,{m^3} at constant pressure 103N/m2,{10^3}N/{m^2}, the work done is A thermodynamic process in which temperature TT of the system remains constant though other variable PP and VV may change, is called What is the slope for an isothermal process? For adiabatic processes (γ=CpCv)\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)
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