A certain volume of gas at 298k is heated such that its volume and pressure are now four times the original values. What is the new temperature?

Answer Details

We can use the ideal gas law to solve this problem, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in kelvin. If the volume and pressure are both increased by a factor of 4, then the new volume V' and new pressure P' are given by: V' = 4V P' = 4P Substituting these values into the ideal gas law, we get: (4P)(4V) = nR(T') Simplifying this equation, we get: 16PV = nRT' Dividing both sides by PV, we get: 16 = nRT' / PV Since n, R, and P are constant, we can simplify this to: 16 = T' / T Solving for T', we get: T' = 16T Therefore, the new temperature is 16 times the original temperature. Substituting T = 298 K, we get: T' = 16 x 298 K = 4768 K So the correct answer is 4768.0K.