Question

Four moles of a gas enclosed in a container is heated at constant pressure. The temperature of the gas increases by \( 3^{\circ} C \). The work done by the gas is \( (R= \) gas constant).

(1) \( 4 R \) 

(2) \( 6 R \) 

(3) \( 8 R \) 

(4) \( 12 R \)

Answer 1

The correct option is (4) 12R.

We have given,

number of moles of gas = 4 mol

change in temperature = 3° c or 3 k

\(\therefore\) work done by gas = -pdv

= -p\(\int^{v2}_{v_1}dv\) (p = constant pressure)

= - p(v₂ - v₁)

= - p \(\left(\cfrac{MRT_2}p-\cfrac{nRT_1}p \right)\) (using ideal gas equation)

= - p x \(\cfrac{nR}p\) (T₂ - T₁)

= - 4 x R x ΔT

= - 4 x R x 3

= - 12 R

Hence, the work done by gas  = 12 R.