Question

a sample of nitrogen gas occupies a volume of 6.90 l at 53°c and 914 mm hg. the volume of the gas sample increases to 8.06 l while its pressure increases to 1.19×10³ mm hg. if the ideal gas law is obeyed, the temperature of the gas sample at the new volume and pressure is °c.

Explanation:

Step1: Convert temperature to Kelvin

The initial temperature \(T_1 = 53^{\circ}C=(53 + 273)K=326K\).

Step2: Apply the combined gas law \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)

We know \(P_1 = 914mmHg\), \(V_1 = 6.90L\), \(P_2=1.19\times10^{3}mmHg\), \(V_2 = 8.06L\).
We need to solve for \(T_2\). Rearranging the combined - gas law formula gives \(T_2=\frac{P_2V_2T_1}{P_1V_1}\).
Substitute the values: \(T_2=\frac{(1.19\times 10^{3})\times8.06\times326}{914\times6.90}\)
First, calculate the numerator: \((1.19\times 10^{3})\times8.06\times326=(1190)\times8.06\times326 = 1190\times2627.56=3.1268\times10^{6}\)
Then, calculate the denominator: \(914\times6.90 = 6306.6\)
\(T_2=\frac{3.1268\times 10^{6}}{6306.6}\approx496K\)

Step3: Convert Kelvin to Celsius

\(T_2=(496 - 273)^{\circ}C = 223^{\circ}C\)

Answer:

\(223\)