Can anyone help me to answer this questions I tried many answers but it is not r

Question

Can anyone help me to answer this questions I tried many answers but it is not rigtht.

A piston contains 220 moles of an ideal monatomic gas that initally has a pressure of 1.41 × 105 Pa and a volume of 2.3 m3. The piston is connected to a hot and cold reservoir and the gas goes through the following quasi-static cycle accepting energy from the hot reservoir and exhausting energy into the cold reservoir. 1. The pressure of the gas is increased to 4.41 x 105 Pa while maintaining a constant volume. 2. The volume of the gas is increased to 9.3 m3 while maintaining a constant pressure. 3. The pressure of the gas is decreased to 1.41 x 105 Pa while maintaining a constant volume. 4. The volume of the gas is decreased to 2.3 m3 while maintaining a constant pressure. It may help you to recall that CV-12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 x 1023) times the number of moles of the gas. 1) How much energy is transferred into the gas from the hot reservoir? 5625606.8 You currently have 3 submissions for this question. Only 10 submission are allowed. You can make 7 more submissions for this question J Submit 2) How much energy is transferred out of the gas into the cold reservoir? J Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 3) How much work is done by the gas? J Submit You currently have O submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 4) What is the efficiency of this cycle?

Explanation / Answer

given, n = 220 moles of gas inside a piston
the gas is monoatomic, so Cp = 5R/2, Cv = 3R/2
initial pressure, P1 = 1.41*10^5 Pa
V1 = 2.3 m^3
using PV = nRT
T1 = P1V1/nR = 177.387 K

step 1 : Pressure increase while maintaining constant volume
   P2 = 4.41*10^5 Pa
   V2 = 2.3 m^3
   T2 = P2V2/nR = 554.808 K

   so Work done by the gas, W1 = 0 ( constant volume case)
   internal energy change , dU, U1 = nCv*(T2 - T1) = 1035001.629 J
   hence heat input to tha gas, Q1 = 1035001.6299 J

step 2 : constant pressure expansion
   P3 = 4.41*10^5 Pa
   V3 = 9.3 m^3
   T3 = P3V3/nR = 2243.354 K

   so work done by gas W2 = P3(VF3 - V2) = 3087000 J
   heat given to the gas, Q2 = nCp(T3 - T2) = 7717499.493 J
   so internal energy change U2 = Q2 - W2 = 4630499.493 J

step 3: COnstant volume pressure decrease
   P4 = 1.41*10^5 Pa
   V4 = 9.3 m^3
   T4 = P4V4/nR = 717.262 K

   so, work done by gas, W3 = 0
   Chnage in internal energy, U3 = nCv(T4 - T3) = -4184999.6742 J
   Hence heat input, Q3 = U3 = -4184999.6742 J

step 4 : constant pressure volume decrease
   P1 = 1.41*10^5 Pa
   V1 = 2.3 m^3
   T1 = 177.387 K

   so, work done by gas = W4 = P1(V1 - V4) = -987000 J
   Heat supplied to the gas, Q4 = nCp(T1 - T4) = -2467498.6875 J
   hence change in internal energy, U4 = Q4 - W4 = -1480498.6875 J

1) Energy tyransferred to the gas from the reservoir, E = Q1 + Q2 = 8752501.1229 J
2) eNERGY TRANSFERRED OUT OF THE RESERVOIR, q = q3 + q4 = 6652498.3617 J
3) Work done by gas W = W1 + W2 + W3 + W4 = 2100000 J
4) efficiency = W/(Q + E) = 0.999999999999999999

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