Let's solve this step-by-step by using the combined gas law:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
where:
- [tex]\( P_1 \)[/tex] and [tex]\( P_2 \)[/tex] are the initial and final pressures,
- [tex]\( V_1 \)[/tex] and [tex]\( V_2 \)[/tex] are the initial and final volumes,
- [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex] are the initial and final temperatures in Kelvin.
### Given:
- Initial temperature ([tex]\(T_1\)[/tex]) = 95°C
- Increase in volume = 25% of the original volume
- Decrease in pressure = one-third of the original pressure
### Step 1: Convert the initial temperature to Kelvin
To work with temperatures in the gas law, we need to convert Celsius to Kelvin. The formula for this is:
[tex]\[ T(\text{K}) = T(\text{C}) + 273.15 \][/tex]
So,
[tex]\[ T_1 = 95 + 273.15 = 368.15 \text{ K} \][/tex]
### Step 2: Calculate the new volume
The volume is increased by 25% of the original volume.
[tex]\[ V_2 = V_1 + 0.25V_1 = 1.25V_1 \][/tex]
Therefore, the ratio of [tex]\( \frac{V_2}{V_1} \)[/tex] is:
[tex]\[ \frac{V_2}{V_1} = 1.25 \][/tex]
### Step 3: Calculate the new pressure
The pressure is decreased by one-third of the original pressure.
[tex]\[ P_2 = P_1 - \frac{1}{3} P_1 = \frac{2}{3} P_1 \][/tex]
Therefore, the ratio of [tex]\( \frac{P_2}{P_1} \)[/tex] is:
[tex]\[ \frac{P_2}{P_1} = \frac{2}{3} \approx 0.6667 \][/tex]
### Step 4: Apply the combined gas law
Since [tex]\( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)[/tex], we can solve for [tex]\( T_2 \)[/tex]:
[tex]\[ T_2 = T_1 \times \frac{V_2}{V_1} \times \frac{P_2}{P_1} \][/tex]
[tex]\[ T_2 = 368.15 \times 1.25 \times 0.6667 \][/tex]
### Step 5: Calculate [tex]\( T_2 \)[/tex]
From the given values, we obtain:
[tex]\[ T_2 \approx 368.15 \times 1.25 \times 0.6667 = 306.7916666666667 \text{ K} \][/tex]
### Step 6: Convert [tex]\( T_2 \)[/tex] back to Celsius
To convert the final temperature back to Celsius:
[tex]\[ T(\text{C}) = T(\text{K}) - 273.15 \][/tex]
[tex]\[ T_2(\text{C}) \approx 306.7916666666667 - 273.15 = 33.64166666666671^\circ \text{C} \][/tex]
### Result:
The new temperature of the gas, when its original temperature is 95°C, the volume is increased by 25%, and the pressure is decreased by one-third, is approximately:
[tex]\[ 33.64^\circ \text{C} \][/tex]
