A certain volume of a gas at 298k is heated such that its volume and pressure are now four times their original value.what is the new temperature?
You know that PV/T is constant. So, you want T' such that
PV/T = (4P)(4V)/T'
T' = (16PVT/PV) = 16T
Now just plug in your value of T.
16
12
We know that PV/T is constant.so,we want T^1 such that PV/t=(4p)(4v)/t^1
T^1=(16PVT/PV)
=16T
So for new temperature,
T^2=initial temperature×volume of gas
=16×298
=4768•0k
To find the new temperature, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature
In this case, we are given that the volume (V) and pressure (P) have increased four times their original value.
Let's assume the original volume and pressure to be V1 and P1 respectively, and the original temperature to be T1.
The new volume (V2) and pressure (P2) would then be:
V2 = 4V1
P2 = 4P1
Now, we need to find the new temperature (T2). Since the amount of gas remains constant, the number of moles (n) will stay the constant as well.
Using the ideal gas law equation, we can rearrange it to solve for temperature:
T = (PV)/(nR)
Substituting the values into the equation, we get:
(T2) = (P2 * V2) / (n * R)
Now, since we are comparing the new temperature (T2) to the original temperature (T1), we can rewrite the equation as:
(T2/T1) = (P2/P1) * (V2/V1)
Plugging in the values we have:
(T2/T1) = (4P1/ P1) * (4V1 / V1)
(T2/T1) = 4 * 4
(T2/T1) = 16
Now, solve for T2:
T2 = T1 * (T2/T1)
Since we are given that the original temperature is 298K, we can substitute T1 = 298K into the equation:
T2 = 298K * 16
T2 = 4768K
Therefore, the new temperature is 4768K.