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A certain mass of gas is expanded from (1L,10Atm) to (4L,5Atm) against a constant external pressure of 1Atm. If the initial temperature of the gas is 300K and the heat capacity of the process is 50JC°. Then the enthalpy change during the process Is: (1LAtm100J)


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Solution

Step 1: Ideal Gas Equation

  1. The Ideal Gas Equation is an equation of state that gives a description about ideal gases.
  2. It relates the pressure, volume, temperature and number of moles of an ideal gas.
  3. 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 m3,

n is the number of moles,

R=8.3145is the universal gas constant in Jmol-1K-1or Pa.m3mol-1K-1,

T is the absolute temperature in K.

Step 2: First Law of Thermodynamics

  1. It is the law of conservation of energy.
  2. 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 moles

Given, for state 1:

V1=1L,P1=10Atm and T1=300K

P1V1=nRT11×10=n×0.0821×300n=0.4

Step 4: Find the value of temperature for state 2

Given, for state 2:

V2=4L,P2=5Atm and n=0.4. Therefore,

P2V2=nRT24×5=0.4×0.0821×T2T2=600K

Step 5: Find the value of change in internal energy

Using first law of thermodynamics, we have

U=Q+WU=50×T+-Pext×VQ=CpTU=50×600-300-1×4-1U=15000-300U=14700J

Step 6: Find the value of enthalpy change

Now we know that enthalpy change, H is given by,

H=U+PVH=14700+5×4-10×1H=14700×10-3+10×100×10-3H=15.7kJ

Hence, the enthalpy change during the process is equal to 15.7kJ.


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