Question

The temperature of an open room of volume $30 \, m^3$ increases from $17^{\circ}C$ to $27^{\circ}C$ due to the sunshine. The atmospheric pressure in the room remains $1 \times 10^5 \, Pa$. If $n_i$ and $n_f$ are the number of molecules in the room before and after heating, then $n_f-n_i$ will be :

• A. $ - 1.61 \times 10^{23}$
• B. $1.38 \times 10^{23}$
• C. $2.5 \times 10^{25}$
• D. $ - 2.5 \times 10^{25}$

Answer 1

The Correct Option is D

To determine the change in the number of molecules, n_f - n_i, as the temperature in a room increases, we can use the ideal gas equation:

PV = nRT

Where:

• P is the pressure,
• V is the volume,
• n is the number of moles of gas,
• R is the universal gas constant (8.314 \, J \, mol^{-1} \, K^{-1}),
• T is the temperature in Kelvin.

Given data:

• Initial temperature T_i = 17^{\circ}C = 17 + 273 = 290 \, K
• Final temperature T_f = 27^{\circ}C = 27 + 273 = 300 \, K
• Volume V = 30 \, m^3
• Pressure P = 1 \times 10^5 \, Pa

Using the ideal gas equation, we have for initial and final states:

n_i = \frac{PV}{RT_i} and n_f = \frac{PV}{RT_f}

To find the change in the number of moles, n_f - n_i:

\begin{align*} n_f - n_i &= \frac{PV}{RT_f} - \frac{PV}{RT_i} \\ &= \frac{PV}{R} \left( \frac{1}{T_f} - \frac{1}{T_i} \right) \\ &= \frac{30 \times 1 \times 10^5}{8.314} \left( \frac{1}{300} - \frac{1}{290} \right) \end{align*}

Calculating the term inside the bracket:

\begin{align*} \frac{1}{300} - \frac{1}{290} &= \frac{290 - 300}{300 \times 290} \\ &= \frac{-10}{87000} \\ &= -\frac{1}{8700} \end{align*}

Substituting back into the equation for n_f - n_i:

\begin{align*} n_f - n_i &= \frac{30 \times 10^5}{8.314} \times \left( -\frac{1}{8700} \right) \\ &= -\frac{30 \times 10^5}{8.314 \times 8700} \\ &= -2.5 \times 10^{25} \end{align*}

Thus, the change in the number of molecules, n_f - n_i, is -2.5 \times 10^{25}.