Suhana is down with fever. So she decided not to go to school tomorrow. Next day, in the evening she called up her classmate, Shaurya and enquired about the computer class. She also requested him to explain the concept. Shaurya said, "Mam taught us how to use tuples in python". Further, he generously said, "Give me some time, I will email you the material which will help you to understand tuples in python". Shaurya quickly downloaded a 2-minute clip from the Internet explaining the concept of tuples in python. Using video editor, he added the text "Prepared by Shaurya" in the downloaded video clip. Then, he emailed the modified video clip to Suhana. This act of Shaurya is an example of: a) Fair use b) Hacking c) Copyright infringement d) Cyber crime

To solve this problem, we'll use the ideal gas law equation $ PV = nRT $.

Given:

• Initial pressure $ P_1 $ = 800 torr
• Initial temperature $ T_1 $ = $0^{\circ}$C = 273.15 K
• Volume $ V $ = 0.50 liters
• Maximum allowable pressure $ P_2 $ = 3.0 atm

First, we need to convert the pressure from torr to atm to keep the units consistent: $$ 1 \text{ atm} = 760 \text{ torr} $$ Hence, $$ P_1 = \frac{800 \text{ torr}}{760 \text{ torr/atm}} = 1.0526 \text{ atm} $$

Now we apply the ideal gas law in the form of $\frac{P_1}{T_1} = \frac{P_2}{T_2}$, where $T_2$ is our unknown temperature in Kelvin.

Solving for $T_2$: $$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$ $$ \frac{1.0526 \text{ atm}}{273.15 \text{ K}} = \frac{3.0 \text{ atm}}{T_2} $$ $$ T_2 = \frac{3.0 \text{ atm} \times 273.15 \text{ K}}{1.0526 \text{ atm}} $$ $$ T_2 \approx 778.50 \text{ K} $$

Convert $T_2$ from Kelvin to Celsius: $$ T_2 (^{\circ}C) = 778.50 \text{ K} - 273.15 \text{ K} $$ $$ T_2 (^{\circ}C) \approx 505.35^{\circ}C $$

So, the highest temperature in Celsius that the vessel can be heated to without exceeding the pressure limit is approximately 505$^{\circ}$C. None of the provided options match exactly 505$^{\circ}$C, but if we consider the nearest value, option 'A' 505 matches most closely.

Thus, the correct answer is:

$$ \text{A) 505} $$

In a glass vessel, a gas is kept at $25^{\circ} \mathrm{C}$ and 0.80 atmosphere pressure. Assuming the vessel can withstand a pressure of 2.0 atmospheres, up to what temperature can this vessel support?

A $745^{\circ} \mathrm{C}$
B $472^{\circ} \mathrm{C}$
C $500^{\circ} \mathrm{C}$
D None of these

Let's solve this problem using the ideal gas law, specifically the relation derived from it when volume and the amount of gas are constant:

$$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$

Where:

• $P_1$ = initial pressure = 0.80 atm
• $T_1$ = initial temperature in Kelvin = $25^{\circ} \mathrm{C} + 273 = 298$ K
• $P_2$ = final pressure = 2.0 atm
• $T_2$ = final temperature in Kelvin

Rearranging for $T_2$:

$$ T_2 = \frac{P_2 \cdot T_1}{P_1} $$

Substitute the given values:

$$ T_2 = \frac{2.0 \cdot 298}{0.80} = 745 \ \text{K} $$

Convert Kelvin back to Celsius:

$$ 745 \ \text{K} - 273 = 472^{\circ} \mathrm{C} $$

Therefore, the temperature the vessel can withstand is

B $472^{\circ} \mathrm{C}$