An ideal gas of molar mass 50 g is given 300 J heat at constant volume. Temperature changes from 20°C to 50°C.

If C_v = 7R/2 and R = 8.3 (SI unit), find mass of gas (in g).

The correct answer is 17.

Step-by-Step Explanation:

First, let us identify the given parameters from the problem statement:

• Heat supplied at constant volume, $Q_v=300\text{J}$
• Molar mass of the gas, $M=50\text{g/mol}$
• Initial temperature, $T_1=20°\text{C}$
• Final temperature, $T_2=50°\text{C}$
• Molar heat capacity at constant volume, $C_v=\frac{7R}{2}$
• Universal gas constant, $R=8.3\text{J/(mol K)}$

Step 1: Calculate the change in temperature
The change in temperature ($\Delta T$) on the Celsius scale is equal to the change in temperature on the Kelvin scale:
$\Delta T=T_2-T_1=50°\text{C}-20°\text{C}=30\text{K}$

Step 2: Relate heat supplied to the number of moles
For a process at constant volume, the heat supplied is related to the number of moles ($n$) and the molar heat capacity ($C_v$) by the formula:
$Q_v=n{C}_v\Delta T$

Step 3: Express the number of moles in terms of mass
The number of moles is given by:
$n=\frac{m}{M}$
where $m$ is the mass of the gas in grams and $M$ is the molar mass in grams per mole.

Substituting this expression into the heat formula gives:
$Q_v=\frac{m}{M}\cdot \left(\frac{7R}{2}\right)\cdot \Delta T$

Step 4: Solve for the mass of the gas ($m$)
Substitute the given values into the equation:
$300=\frac{m}{50}\cdot \left(\frac{7\cdot 8.3}{2}\right)\cdot 30$

Simplify the numerical terms:
$300=m\cdot \left(\frac{30}{50}\right)\cdot \left(\frac{58.1}{2}\right)$
$300=m\cdot 0.6\cdot 29.05$
$300=m\cdot 17.43$
$m=\frac{300}{17.43}\approx 17.21\text{g}$

Rounding to the nearest integer, we find the mass of the gas is approximately 17 g.