Question

A cube 23 cm on each side contains 2.6 g of helium at 20∘C. 1300 J of heat energy are transferred to this gas.

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

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

Answer 1

Get the initial volume and initial pressure for the helium :

V1 = ( 0.23 m )^3 = 0.01217 m^3 = 12.17 L

n = m / M = 2.6 g / 4.003 g per mol = 0.6495 g-moles

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

P1 = ( 0.6495 gmol ) ( 0.08206 atm - L / gmol - K ) ( 293.2 K ) / ( 12.17 L )

P1 = 1.284 atm

For Constant Volume Heat Addition :

T2 = T1 + ( Q ) / ( m ) ( Cv )

T2 = 20 C + ( 1300 J ) / ( 2.6 g ) ( 3.1156 J / g - C )

T2 = 20 C + 160.5 C

T2 = 180.5 C

T2 = 433.5 K

P2 = ( P1 ) ( T2 / T1 )

P2 = ( 1.284 atm ) ( 433.5 K / 293 K ) = 1.899 atm .............Ans(1)

For Constant Pressure Heat Addition :

T2 = T1 + ( Q ) / ( m ) ( Cp )

T2 = 20 C + ( 1300 J ) / ( 2.6 g ) ( 5.1926 J/g - C )

T2 = 116.29 C = 389.29 K

V2 = ( V1 ) ( T2 / T1 )

V2 = ( 12.17 L ) ( 389.29 K / 293 K ) = 16.17 L