What is the equation for the work done by a constant temperature system?  A. W = mRTln(V2-V1)  B. W = mR( T2-T1 ) ln( V2/V1)  C. W = mRTln (V2/V1)  D. W = RT ln (V2/V1) Formula : W=∫ pdV lim1,2 ∫ = mRT / V

What is the equation for the work
done by a constant temperature system?
 A. W = mRTln(V2-V1)
 B. W = mR( T2-T1 ) ln( V2/V1)
 C. W = mRTln (V2/V1)
 D. W = RT ln (V2/V1)
Formula : W=∫ pdV lim1,2
∫ = mRT / V
by

1 Answer

Answer :

W = mRTln (V2/V1)
by

Related questions

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Description : Pick out the Clausius-Clapeyron equation from the following: (A) dP/dT = ∆H/T∆V (B) ln P = - (∆H/RT) + constant (C) ∆F = ∆H + T [∂(∆F)/∂T]P (D) None of these

Last Answer : B) ln P = - (∆H/RT) + constant

The molar excess Gibbs free energy, gE, for a binary liquid mixture at T and P is given by, (gE/RT) = A . x1. x2, where A is a constant. The corresponding equation for ln y1, where y1is the activity co-efficient of component 1, is (A) A . x22 (B) Ax1 (C) Ax2 (D) Ax12

Description : The molar excess Gibbs free energy, gE, for a binary liquid mixture at T and P is given by, (gE/RT) = A . x1. x2, where A is a constant. The corresponding equation for ln y1, where y1is the activity co-efficient of component 1, is (A) A . x22 (B) Ax1 (C) Ax2 (D) Ax12

Last Answer : (A) A . x22

For a perfect gas, according to Boyle’s law (where p = Absolute pressure, v = Volume, and T = Absolute temperature)  A. p v = constant, if T is kept constant  B. v/T = constant, if p is kept constant  C. p/T = constant, if v is kept constant  D. T/p = constant, if v is kept constant

Description : For a perfect gas, according to Boyle’s law (where p = Absolute pressure, v = Volume, and T = Absolute temperature)  A. p v = constant, if T is kept constant  B. v/T = constant, if p is kept constant  C. p/T = constant, if v is kept constant  D. T/p = constant, if v is kept constant

Last Answer : Answer: A

Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1kg of the water?  A. 0.043 K/s  B. 0.179 K/s  C. 0.357 K/s  D. 1.50 K/s Formula: Q = mcv ( T)

Description : Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1kg of the water?  A. 0.043 K/s  B. 0.179 K/s  C. 0.357 K/s  D. 1.50 K/s Formula: Q = mcv ( T)

Last Answer : 0.179 K/s

Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J. Find E. ∆ (Formula: E = Q- W) ∆  a. 125 J  b. 123 J  c. 126 J  d. None of the above

Description : Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J. Find E. ∆ (Formula: E = Q- W) ∆  a. 125 J  b. 123 J  c. 126 J  d. None of the above

Last Answer : 125 J

A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 x10^6 kJ  D. 2 x10^6 kJ Formula: fromthe steamtable vƒ = 0.001043 m^3 / kg vg = 1.6940 m^3 / kg u ƒ= 417.3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Description : A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 ... 3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Last Answer : 2 x10^6 kJ

If w is the angular velocity of the pulley and T1 and T2 are tensions of driving and driven side then power transmitted equals a.(T1 + T2) w b.(T1 + 2T2) w c.107 dynes d.(T1 - T2) w e.wT1

Description : If w is the angular velocity of the pulley and T1 and T2 are tensions of driving and driven side then power transmitted equals a.(T1 + T2) w b.(T1 + 2T2) w c.107 dynes d.(T1 - T2) w e.wT1

Last Answer : d. (T1 - T2) w

A gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04 m3 to 0.10 m3 at a constant pressure of 200 kPa. Find the work done on the system.  a. 5 kJ  b. 15 kJ  c. 10 kJ  d. 12 kJ

Description : A gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04 m3 to 0.10 m3 at a constant pressure of 200 kPa. Find the work done on the system.  a. 5 kJ  b. 15 kJ  c. 10 kJ  d. 12 kJ

Last Answer : 12 kJ

According to first law of thermodynamics  (a) work done by a system is equal to heat transferred by the system  (b) total internal energy of a system during a process remains constant  (c) internal energy, enthalpy and entropy during a process remain constant  (d) total energy of a system remains constant

Description : According to first law of thermodynamics  (a) work done by a system is equal to heat transferred by the system  (b) total internal energy of a system during a process remains constant  ( ... , enthalpy and entropy during a process remain constant  (d) total energy of a system remains constant

Last Answer : Answer : d

Isochoric process is one in which  (a) free expansion takes place  (b) very little mechanical work is done by the system  (c) no mechanical work is done by the system  (d) all parameters remain constant

Description : Isochoric process is one in which  (a) free expansion takes place  (b) very little mechanical work is done by the system  (c) no mechanical work is done by the system  (d) all parameters remain constant

Last Answer : Answer : c

In a non-flow reversible process for which p = (- 3V+ 15) x 105N/m2,V changes from 1 m to 2 m3. The work done will be about  (a) 100 xlOO5 joules  (b) lxlO5 joules  (c) 10 xlO5 joules  (d) 10 xlO5 kilo joules  (e) 10xl04ki\ojoules.

Description : In a non-flow reversible process for which p = (- 3V+ 15) x 105N/m2,V changes from 1 m to 2 m3. The work done will be about  (a) 100 xlOO5 joules  (b) lxlO5 joules  (c) 10 xlO5 joules  (d) 10 xlO5 kilo joules  (e) 10xl04ki\ojoules.

Last Answer : Answer : c