Question

30.0 grams of ideal CO2 at 310 K and 1 atm expands until the pressure is halved. The temperature is unchanged. Find q for this process.

Answer 1

According to the first law of thermodynamics change in the internal energy of a system is given by

ΔU = q + w

where,

ΔU = Change in the internal energy of the system

q = Heat added to the system

w = Work done

We are given that the temperature remains constant. Thus, the given process is a isothermal process.

We know that for a isothermal process, ΔU = 0

Thus, q = w   (work done is positive beacuse the gas is expanding)

Work done in a isothermal process (w) = nRT ln(P₁ / P₂)

Note : Here ln is the natural log

where,

n = No. of moles of gas

R = Gas constant

T = Temperature of gas

P₁ = Initial Pressure

P₂ = Final Pressure

We are given,

R = 8.314 J mol^-1 K^-1

T = 310 K

P₁ = 1 atm

P₂ = Half of the initial pressure = 0.5 atm

First calculate the no. of moles of CO₂ present,

Molar mass of CO₂ = 44 g

Mass of CO₂ given = 30 g

No. of moles (n) of CO₂ = Mass of CO₂ given / Molar Mass of CO₂

No. of moles (n) of CO₂ = 30 / 44 = 0.682 mol

On substituting the values in the equation we get,

Work done (w) = (0.682 mol) * (8.314 J mol^-1 K^-1) * (310 K) * ln(1 atm / 0.5 atm)

Work done (w) = 1218.376 J                                 (ln (2) = 0.693)

As we know,

q = w

Thus, q = 1218.376 J