Question

A cube 23 cm on each side contains 2.8 g of helium at 20∘C. 1300 J of heat energy are transferred to this gas.1-What is the final pressure if the process is at constant volume? 2-What is the final volume if the process is at constant pressure?

Answer 1

Volume of cube V1 = ( 0.23 m )^3
= 0.01217 m^3
= 12.17 L

n = m / M
= 2.8 g / 4.003 g per mol
= 0.6995 gmoles

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

P1 = ( 0.6995 * 0.08206 * 293.2 ) / ( 12.17 )

P1 = 1.383 atm

For Constant Volume Heat Addition :

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

T2 = 20 C + ( 1300 J ) / ( 2.8 g ) ( 3.1156 J / g - C )

T2 = 169.02 C
= 442.02 K

P2 = ( P1 ) ( T2 / T1 )

P2 = ( 1.383 atm ) ( 442.02 K / 293 K )

= 2.086 atm

b)

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

T2 = 20.0 C + ( 1300 J ) / ( 2.8 g ) ( 5.1926 J/g - C )

T2 = 109.413 C
= 382.413 K

V2 = ( V1 ) ( T2 / T1 )

V2 = ( 12.17 L ) ( 382.413 K / 293.2 K )
= 15.9 L