• Calculate total work done in an adiabatic process.

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A process, in which the working substance neither receives nor gives out heat to its surroundings during its expansion or compression, is called: (VIZAG MT Mech 2015)

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Work done in an adiabatic process between a given pair of end states depends on: (CIL MT Mech 2020)

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An isolated system is one in which:

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The ratio of specific heat at constant pressure (C$_p$) and specific heat at constant volume (C$_v$) is:

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Characteristic gas constant of a gas is equal to: (ISRO Scientist ME 2010)

Solution:
Characteristics Gas Constant, $R = C_p - C_v$

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The gas constant (R) is equal to the _________ of two specific heats. (SSC JE ME Paper 3- Sep 2019 Evening)

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The value of gas constant (R) in S. I. units is: (RRC Group D Oct 2018 Shift 3)

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When a perfect gas is heated at constant pressure from 15 $^\circ$C to 95$^\circ$ C, the heat required is 1136kJ/kg. When the same gas is heated at constant volume between the same temperatures the heat required is 808 kJ/kg. Calculate $\gamma$ for the gas.

Solution:
Constant Volume Process $Q = 808\ kJ/kg$ $T_1 = 15 + 273 = 288\ K,$ $T_2 = 95 + 293 = 368\ K,$ $\Delta T = T_2 - T_1 = 80\ K$ $Q = mC_v \Delta T, 806 = 1 \times C_v \times 80, C_v = 10.1\ KJ/kg K$ Constant Pressure Process \begin{align*} Q &= 1136\ KJ/kg\\ Q &= m C_p \Delta T\\ 1136 &= 1 \times C_p \times 80\\ C_p &= 14.2\ kJ/kg K \end{align*} $\gamma = \dfrac{C_p}{C_v} = \dfrac{14.2}{10.1} = 1.405$

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A mono-atomic ideal gas ($\gamma$ = 1.67, molecular weight = 40 ) is compressed adiabatically from 0.1 MPa , 300 K to 0.2 MPa . The universal gas constant is 8.314 kJ/KmolK. The work of compression of the gas is: (GATE ME 2010)

Solution:
Characteristic Gas Constant, $r = \dfrac{R}{M} = \dfrac{8.314}{40} = 0.208 kJ/kgK$ Using, $\left(\dfrac{T_2}{T_1}\right) = \left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma - 1}{\gamma}}$ $T_2 = 396.18K$ Work done in compression, $= \dfrac{r(T_2 - T_1)}{\gamma - 1} = \dfrac{0.208 (396.18 - 300)}{1.67 - 1} = 29.8$

5m$^3$ of air at 2 bar and 27$^\circ$C is compressed upto 6 bar. This process is governed by equation pV$^{1.3}$ = C. It is subsequently expanded to 2 bar. This process is reversible adiabatic. Find the net work in kJ: