Question

Suppose that 2 mol of gas is measured at 2 atm pressure to have a volume of 5 L, a temperature of 300 K. If the conditions change such that there are 4 mol at 4 atm and at 270 K, what is the new volume?

Answer 1

Q97088

Solution:

Given:

Initial conditions;

$n_i=2 mol$ ---(initial no. of moles)

$P_i=2 atm.$ ---(initial pressure)

$V_i=5 L$ ---(initial volume)

$T_i=300K$ ---(initial temperature of gas)

Final conditions;

$n_f=4 mol$ ---(final no. of moles)

$P_f=4 atm.$ ---(final pressure)

$T_f=270K$ ---(final temperature)

To find: $V_f$ (final volume of the gas)

Using Ideal Gas Law; $PV=nRT$

where, $R=gas constant$

$\therefore PV/nT=constant$

$\implies P_iV_i/n_iT_i= P_fV_f/n_fT_f$

$\therefore V_f=(P_iV_i/n_iT_i)*(n_fT_f/P_f)$

Substituting the given values, we get;

$V_f=((2atm*5L)/(2 mol*300K))$ $*(4mol*270K/4 atm)$

$=4.5L$ (Answer)

Thus new volume of the gas is $4.5 L$