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# A certain mass of gas is expanded from $\left(1L,10Atm\right)$ to $\left(4L,5Atm\right)$ against a constant external pressure of $1Atm$. If the initial temperature of the gas is $300K$ and the heat capacity of the process is $50J}{{}^{°}C}$. Then the enthalpy change during the process Is: $\left(1LAtm\simeq 100J\right)$

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Solution

## Step 1: Ideal Gas Equation The Ideal Gas Equation is an equation of state that gives a description about ideal gases.It relates the pressure, volume, temperature and number of moles of an ideal gas.It is a combination of Boyle's law, Charles' law and Avogadro's law.Mathematically, Ideal Gas Equation is given by$PV=nRT$, Where$P$ is the pressure in $Pa$,$V$ is the volume in ${m}^{3}$,$n$ is the number of moles,$R=8.3145$is the universal gas constant in $Jmo{l}^{-1}{K}^{-1}$or $Pa.{m}^{3}mo{l}^{-1}{K}^{-1}$,$T$ is the absolute temperature in $K$.Step 2: First Law of ThermodynamicsIt is the law of conservation of energy.It states that the amount of energy added to the system in the form of heat is equal to the sum of the change in internal energy of the system and the energy leaving the system in the form of work done by the system on its surroundings.Mathematically, the first law of thermodynamics is given by$Q=∆U+W$,where$Q$ is the amount of the heat energy absorbed by the system,$W$ is the work done by the system on its surroundings,$∆U$ is the change in internal energy of the system.Step 3: Find the value of number of molesGiven, for state $1$:${V}_{1}=1L,{P}_{1}=10Atm$ and ${T}_{1}=300K$${P}_{1}{V}_{1}=nR{T}_{1}\phantom{\rule{0ex}{0ex}}⇒1×10=n×0.0821×300\phantom{\rule{0ex}{0ex}}⇒n=0.4$Step 4: Find the value of temperature for state 2Given, for state $2$:${V}_{2}=4L,{P}_{2}=5Atm$ and $n=0.4$. Therefore,${P}_{2}{V}_{2}=nR{T}_{2}\phantom{\rule{0ex}{0ex}}⇒4×5=0.4×0.0821×{T}_{2}\phantom{\rule{0ex}{0ex}}⇒{T}_{2}=600K$Step 5: Find the value of change in internal energyUsing first law of thermodynamics, we have$∆U=Q+W\phantom{\rule{0ex}{0ex}}⇒∆U=\left(50×∆T\right)+\left(-{P}_{ext}\right)×∆V\left[\because Q={C}_{p}∆T\right]\phantom{\rule{0ex}{0ex}}⇒∆U=50×\left(600-300\right)-\left(1×\left(4-1\right)\right)\phantom{\rule{0ex}{0ex}}⇒∆U=15000-300\phantom{\rule{0ex}{0ex}}⇒∆U=14700J$Step 6: Find the value of enthalpy changeNow we know that enthalpy change, $∆H$ is given by,$∆H=∆U+∆\left(PV\right)\phantom{\rule{0ex}{0ex}}⇒∆H=14700+\left(\left(5×4\right)-\left(10×1\right)\right)\phantom{\rule{0ex}{0ex}}⇒∆H=14700×{10}^{-3}+10×100×{10}^{-3}\phantom{\rule{0ex}{0ex}}⇒∆H=15.7kJ$Hence, the enthalpy change during the process is equal to $15.7kJ$.

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