The temperature of a 2.5 L of a gas intially at STP is raised to 523K at constant volume.?

Question

Calculate the final pressure of gas in atm.

Answer 1

okay...this is most def a gas law problem...

at STP p=1 atm, and temperature=273 K

therefore, using this identity:

(P1 * V1)/ T1 = (P2 * V2)/ T2,

we can calculate the final pressure of the gas in atm by plugging in the appropriate information given to us...:

P1= 1 atm
V1= 2.5 L
T1= 273 K

P2=what we are solving for "x"
V2= 2.5 L (volume is constant)
T2=523

therefore, since V is constant, V1=V2, and the "volumes" in the above identity cancel and we are left with

P1/T1 = P2/T2

(1 atm)/(273 K) = (P2)/(523)

by cross multiplying and dividing, we get:

273*P2=523
P2=1.915 (rounded to 3 decimal places)

Answer 2

You can use P1*V1/T1=P2*V2/T2 to solve this. Since the problem says "constant volume", V1 and V2 will be the same, so we don't need to consider them. So the equation now becomes P1/T1=P2/T2. We are looking for P2, the final pressure, so P1, T1, and T2 should be given to us in the equation. The initial conditions are at STP, or Standard Temperature and Pressure, which is 273K and 1 atm (just something you have to know or can look up). So P1=1 T1=273 and T2=523, also from the question. So we get:

1/273=x/523

Now just solve for x. I'll let you do this. =)

Answer 3

the growth cuts the rigidity from 1515 kPa to 505 kPa, so it is going to triple the quantity of the cylinder simply by fact the temperature remains consistent (it fairly is isothermal). V?/V? = 3. additionally, simply by fact the growth is isothermal, ?E = 0 for the gas (the interior power relies upon basically on temperature, assuming the gas may well be dealt with as suitable). as a result, the artwork carried out on the encircling could be balanced by the warmth that is going into the gas interior the cylinder, w = -q The artwork on our environment is basically w(surr) = -w(sys) = P(ext)?V = (505 kPa)(V? – V?) = 2(505 kPa)(V?), considering that V? = 3V?. ?S(surr) = q(surr)/T(surr) = –w(surr)/T = –2(505 kPa)(V?)/T = –2(505 kPa)(nRT/P?)/T ?S(surr) = –2(505 kPa)(nR)/(1515 kPa) = –(?)nR All you're able to do is artwork out n using the suitable gas regulation, and additionally you will have your answer.

Answer 4

P2V2/T2 = P1V1/T1
V1 = V2
P2 = P1T2/T1
P2 = (1 atm)(523/293)
P2 = 5.6237 atmospheres