The temperature of the gas is 41.3 °C.
Answer:
The temperature of the gas is 41.3 °C.
Explanation:
So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923×
atm. Thus, 171000 Pa = 1.6876 atm.
We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.
Then substituting all the values in ideal gas equation ,we get
1.6876×50=3.27×0.0821×T
Temperature = 
So the temperature is obtained to be 314.3 K. As 0°C = 273 K,
Then 314.3 K = 314.3-273 °C=41.3 °C.
Thus, the temperature is 41.3 °C.
Answer:
If the mass of B is m and the temperature change is the same, the mass of B will be 2m.
Explanation:
Q = mcT
T = mc/Q
M = 4Q/2cT........... (1)
T = Q/mc
Plug this in equation 1.
M = 4Q/(2c × Q/mc) = 4Q ÷ 2Q/m = 4Q × m/2Q = 2m
Answer: B
Explanation:
Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.
Answer:
24m/s²
Explanation:
Given
Distance S = 3m
Time of fall = 0.5sec
Required
Acceleration due to gravity
Using the equation of motion
S = ut+1/2gt²
Substitute the given values
3 = 0+1/2g(0.5)²
3 = 1/2(0.25)g
3 = 0.125g
g = 3/0.125
g = 24
Hence the value for the acceleration of gravity on this new planet is 24m/s²
20/40=0.5 g/cm^3 becuase, mass/volume=density.
