Question

2. A cylinder closed by a moveable piston, contains an ideal gas at 50℃ at atmospheric pressure 1 X 10 4Pa. The piston is moved out until the volume of gas is doubled and the temperature falls to 25℃. The new pressure of the gas in the cylinder is​

Answer 1

Explanation:

Q

1

=dU+dW=nc

v

dT+PdW=nc

v

(70)+PAx

x=0.2m

When the piston is fixed dV=0 therefore workdone is zero

Q

2

=dU=nc

v

dT=nc

v

(70)

Difference of heat= PAx=(P

atm

+mg/A)Ax=136J

Answer 2

Explanation:

acc to the question the gas is ideal gas so it will follow the ideal gas equation,

GIVEN,

Initial temperature = 50° C

Initial pressure = 1× $10^{4}$ Pa

let the initial volume = $V_{1}$

after the movement of piston,

temperature = 25°C

Let the new pressure = $P_{2}$

Let the volume =$2V_{1}$

Now using ideal gas equation,

$P_{1}$ ×$V_{1} /T_{1}$ =$P_{2}$× $V_{2} /T_{2}$

1×10×V /50 = $P_{2}$ × 2V /25

1×10×V×25=$P_{2}$

 50×2V

from here we can calculate the pressure,

$P_{2}$ = 1×$10^{4}$/4

$P_{2}$ = $0.25$ ×$10^{4}$ Pa

so here we get the pressure.