Question

A cube 22 cm on each side contains 3.4 g of helium at 20∘C. 1200 J of heat energy are transferred to this gas.

Part A What is the final pressure if the process is at constant volume?

p =

Part B What is the final volume if the process is at constant pressure?

V =

Answer 1

Get the initial volume and initial pressure for the helium :

V1 = ( 0.22 m )^3 = 0.010648 m^3 = 10.648 L

n = m / M = 3.4 g / 4.003 g per mol = 0.8493 gmoles

P1 = ( n ) ( R ) ( T1 ) / ( V1 )

P1 = ( 0.8493 gmol ) ( 0.08206 atm - L / gmol - K ) ( 293.2 K ) / ( 10.648 L )

P1 = 1.919 atm

For Constant Volume Heat Addition :
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T2 = T1 + ( Q ) / ( m ) ( Cv )

T2 = 20 C + ( 1200 J ) / ( 3.4 g ) ( 3.1156 J / g - C )

T2 = 20 C + 168.1 C = 133.28 C = 406.2K

P2 = ( P1 ) ( T2 / T1 )

P2 = ( 1.919 atm ) ( 406.2 K / 293.2 K ) = 2.659 atm <------------

For Constant Pressure Heat Addition :
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T2 = T1 + ( Q ) / ( m ) ( Cp )

T2 = 20.0 C + ( 1200 J ) / ( 3.4g ) ( 5.1926 J/g - C )

T2 = 87.97 C = 361.1K

V2 = ( V1 ) ( T2 / T1 )

V2 = ( 10.648 L ) ( 361.1 K / 293.2 K ) = 13.11 L <--------------------------