Answer to Question #319154 in Physical Chemistry for HRMSCA

Question

Kindly calculate ΔE , the q and w for the following processes of an ideal gas going from State 1 to State 2.

State 1

a) P= 2.50 atm V=8.97 L T= 273.15 K

b) P= 2.50 atm V=15.5 L T= 473.15 K

c) P= 1.45 atm V=15.5 L T= 273.15 K

State 2

a) P= 2.50 atm V=15.5 L T= 473.15 K

b) P= 1.45 atm V=15.5 L T= 273.15 K

c) P= 2.50 atm V=8.97 L T= 273.15 K

Answer 1

If an ideal gas is a monoatomic gas then C_v = 3/2 * R;

R = 0.082 L * atm/(K * mol);

a)

State 1: P= 2.50 atm V=8.97 L T= 273.15 K;

State 2: P= 2.50 atm V=15.5 L T= 473.15 K;

This is the isobaric process, then:

W = P * ΔV = 2.5 * (15.5 - 8.97) = 2.5 * 6.53 = 16.3;

ΔE = C_v * n * ΔT = 3/2 * 0.082 * 1 * (473.15 - 273.15) = 24.6;

Q = ΔE + W = 24.6 + 16.3 = 40.9.

b)

State 1: P= 2.50 atm V=15.5 L T= 473.15 K;

State 2: P= 1.45 atm V=15.5 L T= 273.15 K;

This is the isochoric process, then:

W = 0 (an ideal gas doesn't do any work);

ΔE = Q = C_v * n * ΔT = 3/2 * 0.082 * 1 * (273.15 - 473.15) = -24.6;

Q = ΔE = -24.6.

c)

State 1: P= 1.45 atm V=15.5 L T= 273.15 K;

State 2: c) P= 2.50 atm V=8.97 L T= 273.15 K;

This is the isothermal process, then:

W = n * R * T * ln(V₂/V₁) = 1 * 0.082 * 273.15 * ln(2.50/1.45) = 12.2;

ΔE = 0 (initial energy doesn't change);

Q = W = 12.2.