JEE PYQ: Thermodynamics Question 4
Question 4 - 2021 (25 Feb Shift 2)
Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external 4.3 MPa. The heat transferred in this process is ___ kJ mol$^{-1}$. (Rounded-off to the nearest integer) [Use $R = 8.314$ J mol$^{-1}$ K$^{-1}$]
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Answer: 15
Solution
Moles $(n) = 5$, $T = 293$ K. Process = Isothermal, Irreversible. $P_{ini} = 2.1$ MPa, $P_f = 1.3$ MPa, $P_{ext} = 4.3$ mPa. Work $= -P_{ext}\Delta v = -4.3 \times \left(\dfrac{5 \times 293R}{1.3} - \dfrac{5 \times 293R}{2.1}\right) = -5 \times 293 \times 8.314 \times 43\left(\dfrac{1}{13} - \dfrac{1}{21}\right)$. $\approx -15347$ J $= -15.34$ kJ. Isothermal process, so $\Delta U = 0$, $w = -Q$. $Q = 15.34$ kJ/mol. Answer is 15.
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