The volume of some air assumed to be an ideal gas in the...

The volume of some air, assumed to be an ideal gas, in the cylinder of a car engine is 540 cm^3 at a pressure of 1.1 x 10^5 Pa and a temperature of 27 degrees Celsius. The air is suddenly compressed, so that no thermal energy enters or leaves the gas, to a volume of 30 cm^3. The pressure rises to 6.5 x 10^6 Pa. Determine the temperature of the gas after compression.

Transcript

00:01 Hello students over a question given to us the volume of some air we are assuming it to be an ideal gaze and we are compressing in and after the comparison we find out the temperature of the case after the comparison so let's see in we are looking first the given quantities which are initial pressure we are showing it back even that is 1 .1 this is given in the quotient multiply by 10 to the power 5 pascal because it's given 1 atm okay then initial volume b1 which is given 540 centimeter cube so we are converting in meter cube that is 540 multiply by 10 to the power minus 3 so the initial temperature given to us is 27 degrees celsius so we are converting in it kelvin so we are just adding to 73 here so this will be 350 .15 .15 kelvin and the final pressure p2 after the compression okay this given to us 6 .5 multiply by 10 to the power 6 pascal and the volume final volume we two given to us is 30 centimeter cube so we are converting here it in meter cube we are multiplying by 10 to the power minus 3 so this is meter cube and the we have to find out the trampleger that after the compression what will be the temperature of the gays so so as they are given that there is no energy exchange between the air and the surrounding so this is clear that this process is adiabatic this is the adiabatic process because there is no energy exchange that means that means no heat exchange okay so therefore we can apply the formula this is the formula when there is an adiabatic process so this is p v to the power gamma is equal to constant so we are giving it the equation one so as p v to the power gamma is how that is as equals two concept so we can write down it like this p1 v1 to the power gamma and p2 v to the power gamma where gamma is the capacity ratio that is cp by cv so we are putting here the given quantity is the value of b1 that is 1 .1 multiply by 10 to the power 5 the value of v1 that is 5 440 multiplied by 10 to the power 5 the value of v1 to the power minus 3 and its power is gamma is equal to 6 .5 multiply by 10 to the power 5 and multiply by the value of v2 is 30 multiplied by 10 to the power minus 3 and its power is so we are doing it here okay after this we are doing here so after this we will get 540 by 30 to the power gamma and is equal to 6 .5 multiply 10 to the power 6 this is 6 okay 10 to the power 6 and multiply by 1 and divide by 1 .1 multiply by 10 to the power 5.

03:07 So how it will come? 10 to the power minus 3 and 0 to the power gamma and 10 to the power minus 3 to the power gamma will be cancelled out.

03:14 5 .40 this is 30 because their powers are same and we will do like this 540 by 30 and it's power gamma.

03:21 And this is 6 .5 multiplied by 10 to power 6.

03:24 So what we are doing here we are putting it in denominator.

03:26 So this is, so this will be 540 divided by 30 and the power gamma is equals to 650 divided by 11.

03:36 This is equation 2...