Question

The temperature of 20 L of $N_2$ was increased from 100 K to 300 K at a constant pressure change in volume:

• A. 80 L
• B. 60 L
• C. 40 L
• D. 20 L

Answer 1

The correct option is C 40 L

Given,

Initial temperature $T_1$=100K

Final temperature $T_2$=300K

Volume initially $V_1$=20L

Also pressure is constant

We know by charle's law at constant P

$V\propto T$

$\frac{V_1}{V_2}=\frac{T_1}{T_2}$

$V_2$$=300\times \frac{20}{100}=60L$

Volume finally =60L

New change is volume =$V_{final}-V_{initial}$

$60-20=40L$