Question

The volume of a gas measured at $27^{\circ} C$ and $1 \,atm$ pressure is $10\, L$. To reduce the volume to $2 \,L$ at $1\, atm$ pressure, the temperature required is

• A. 60 K
• B. 75 K
• C. 150 K
• D. 225 K

Answer 1

Answer: (A)

60 K

Answer 2

Here $V_{1} =10\, L , V_{2}=2\, L$ $p_{1} =1 \,atm , p_{2}=1 \,atm$ $T_{1}=300\, K , T_{2}=?$ $\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}}$ $\frac{1 \times 10}{300}=\frac{1 \times 2}{T_{2}}$ $T_{2}=\frac{300 \times 2}{10}=60 \,K$