A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6.0m^3. How much work is done by the gas if it expands
at constant pressure to twice its initial volume?
Please help me I really don't know what to do,,,
To find the work done by the gas, you can use the formula:
Work = Pressure * Change in Volume
In this case, the pressure is given as 1.3x10^5 Pa and the gas expands to twice its initial volume, which means the change in volume is:
Change in Volume = Final Volume - Initial Volume
= 2 * Initial Volume - Initial Volume
= Initial Volume
Given that the initial volume is 6.0 m^3, the change in volume is 6.0 m^3.
Now, you can substitute the values into the formula for work:
Work = Pressure * Change in Volume
= (1.3x10^5 Pa) * (6.0 m^3)
= 7.8x10^5 Pa*m^3
Therefore, the work done by the gas when it expands at constant pressure to twice its initial volume is 7.8x10^5 Pa*m^3.