Question

$4l$ of a gas at ${27}^{o}C$ and $1$ atm is heated to ${227}^{o}C$ and $2$ atm. The new volume under the above condition is:

• A. $4l$
• B. $3l$
• C. $\frac{10}{3}l$
• D. $\frac{11}{3}l$

Answer 1

Answer: (C)

$\frac{10}{3}l$

From ideal gas equation-

$PV=nRT$

For the same gas with same no. of moles,

$\frac{PV}{T}=\text{constant}$

$\Rightarrow \frac{P_1{V}_1}{P_2{V}_2}=\frac{T_1}{T_2}.....\left(1\right)$

From ${eq}^n\left(1\right)$, we have:

$\frac{1\times 4}{2\times V_2}=\frac{300}{500}$

$\Rightarrow V_2=\frac{10}{3}L$

Hence the new volume under the given conditions is $\frac{10}{3}L$.

Hence, the correct answer is option $\text{C}$.