A cylinder that contains methane gas is kept at a temperature of 15°C and exert a pressure of 7atm.If the temperature of the cylinder increases to 25°C,what pressure does the gas now exert?

I don't think we have an tutors in the School Subject "Ellison". I've never heard of it.

since V is constant, and PV=kT,

P/T = k/V is constant. So, you want P such that

P/(273+25) = 7/(273+15)

To calculate the new pressure exerted by the gas when the temperature increases, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature of the gas

Since we are keeping the volume constant, we can rewrite the equation as:

P1/T1 = P2/T2

Where:
P1 = initial pressure
T1 = initial temperature
P2 = final pressure (what we need to find)
T2 = final temperature

Given:
P1 = 7 atm (initial pressure)
T1 = 15°C = 15 + 273 = 288K (initial temperature)
T2 = 25°C = 25 + 273 = 298K (final temperature)

Now we can substitute the values into the equation:

7 atm / 288K = P2 / 298K

To solve for P2, we can cross-multiply:

P2 = (7 atm * 298K) / 288K

Calculating this equation will give us the final pressure:

P2 ≈ 7.24 atm

Therefore, when the temperature of the cylinder increases from 15°C to 25°C, the methane gas will exert a pressure of approximately 7.24 atm.