this is for pchem i need help solving 3 to 8 3. One mol of He gas with Rexpands

Question

this is for pchem
i need help solving 3 to 8

3. One mol of He gas with Rexpands reversible from 24.6 L and 300 K to 40.2 L Calculate the final pressure and temperature if the expansion is: a) Isothermal. b) Adiabatic. Derive an equation for U for an ideal gas that is heated from T, to Ta and the constant volume heat capacity is given by Cya+bT+CT where a, b, and c are constant. What are the units of a, b, and c in (J, K, and )? 4, 5. One mole of an ideal gas initially at 27°C and 1 atm pressure is heated and allowed to expand reversibly at constant pressure until the final temperature is 327°c a) Calculate w b) Calculate AU and AH if CR and CR c) Calculate q 6. The molar heat capacity of liquid water C.-75.4 heated from room temperature to boiling point. Calculate AU if s00 g of liquid water is mol - K ree moles of an ideal gas initially at 280 K is expanded from an initial volume of 13.0 L to a final volume of 30.0 L. Calculate au, q, w, AH, and the final temperature T2 for this expansion carried out according to each of the following path: a) Isothermal, reversible expansion b) Expansion against an external pressure of 1.0 atm under adiabatic condition. c) An expansion against a zero external pressure (vacuum) in an adiabatic system. 8. For each of the following processes deduce whether each of the quantities: q, w, AU, and AH are positive, negative or o. a) Reversible melting of solid benzene at 1 atm and the normal melting temperature. b) Reversible adiabatic expansion of a perfect gas c) Reversible isothermal expansion of perfect gas. d) Adiabatic expansion of ideal gas into vacuum.

Explanation / Answer

gas law can be written as

P1V1/T1= P2V2/T2, where P1,V1,T1 refer to conditions 1 and P2,V2 and T2 refer to conditions 2.

At constant temperature, the equation becomes

P1V1= P2V2 given P1 need to be determined from gas law PV= nRT, P1= nRT/V1

P1V1 becomes nRT=P2V2

Given n=1 , R=0.0821 L.atm/mole.K, T= 300K, V2= 40.2L

Hence P2= 1*0.0821*300/40.2 =0.61 atm, since the expansion is isothermal, the temperature remains at 300K.

For adiabatic process, P1V1Y= P2VY   (1), Y= ratio of specific heats =CP/CV= (CV+R)/CV= (1.5R+R)1.5R= 1.67

P1= nRT/V1= 1*0.0821*300/24.6= 1atm, from Eq.1, P2= P1*(V1/V2)Y= 1*(24.6/40.2)1.67 = 0.44 atm

From gas law P1V1/T1= P2V2/T2, T2= P2V2T1/P1V1= 0.44*40.2*300/(1*40.6)= 131 K

,

3.

For constant pressure process,

Work done = -n*R*(T2-T1)

Where n = no of moles= 1

Hence work done = -8.314*(327+273-(27+273)

=-8.314*300 joules=-2494.2 joules

deltaH=Q=nCp*deltaT= 1*2.5*8.314*300=6235.5 joules

deltaU= nCV*deltaT= 1*1.5*8.314*300 =3741.3 joules

4.

Mass of water= 500 gm, converting this into moles, moles= mass/molar mass= 500/18 =27.8 moles

deltaU= nCv*deltaT, deltaT= 100( boiling point)- 27 ( room temperature)= 73 deg.c

and n= no of moles. CV= 75.4 J/mole.K

deltaU= 27.8*75.4*73 joules=153017 joules

5.

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