Question

A cube 17 cm on each side contains 2.4 g of helium at 20∘C. 1400 J of heat energy are transferred to this gas.

1.)What is the final pressure if the process is at constant volume?(in atm)

2.)What is the final volume if the process is at constant pressure?(in cm^3)

Answer 1

a = 17 cm = 0.17m; m = 2.4 g ; t = 20 deg C = 293 K; E = 1400 J

Volume is:

V = a^3 = 0.17^3 = 4.91 Lit

n = m/M = 2.4/4.003 = 0.599 gmol

Pressure will be:

P1 = n R T1 / V1

P1 = 0.599 x 0.08206 x 293/4.91 = 2.933 atm

A)T2 = T1 + E/m Cv

T2 = 20 + 1400/2.4 x 3.1156 = 207.23 deg C = 480.23 k

Now using P1V1/T1 = P1V1/T2

at constant volume,

P2 = P1 (T2/T1)

P2 = 2.933 atm (480.23/293) = 4.81 atm

Hence, P2 = 4.81 atm

B)temp now will be:

T3 = T1 + E/m Cp

T3 = 20 + 1400/2.4 x 5.1926 = 132.34 = 405.34 K

Now using P1V1/T1 = P1V1/T2

at constant Pressure,

V2 = V1 (T2/T1)

V2 = 4.91 lit (353.82/293) = 6.8 lit

Hence, V2 = 6.8 lit = 6800 cm^3.