Answer
$0.892\ m^3$
Work Step by Step
Given;
Temperature of ideal gas is $T_1 =10^{\circ} C $
We then convert given temperature from degrees to kelvins i,e $10^{\circ} C+273 = 283 K $
Pressure $P_1 = 100\ kpa = 100\times 10^3 pa = 10^5 pa$
Volume $V_1 =2.50\ m^3$
Pressure raised to $P_2 =300\ kpa =300\times 10^3 pa $
Temperature increases to $T_2 = 30^{\circ} C =30^{\circ} C+273 =303 K$
Let the volume occupied by gas be $V_2$
From ideal gas law we have the formula;
$PV = nRT$
From this, we find that $PV\propto T$.
So, we can write;
$\frac{P_1 V_1}{T_1}=\frac{P_2V_2}{T_2}$
$V_2 =\frac{P_1V_1T_2}{P_2T_1}$
$V_2 =\frac{(10^5)(2.5)(303)}{(300\times 10^3)(283)}$
$V_2 =0.892\ m^3$
